Changed layout for more manageable volumes

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-#LyX 2.4 created this file. For more info see https://www.lyx.org/
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-\backslash
-exerc1[00]
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-\end_inset
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-How many combinations of 
-\begin_inset Formula $n$
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-\end_inset
-
-
-\begin_inset Formula $\binom{n}{n-1}=\frac{n!}{1!(n-1)!}=n$
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-.
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-\backslash
-exerc3[00]
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-
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-How many bridge hands (13 cards out of a 52-card deck) are possible?
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-
-\begin_inset Formula $\binom{52}{13}=635013559600$
-\end_inset
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-.
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-\backslash
-exerc4[10]
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-Give the answer to exercise 3 as a product of prime numbers.
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-\backslash
-answer 
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-
-\end_inset
-
-We have 
-\begin_inset Formula $\binom{52}{13}=\frac{52!}{13!39!}$
-\end_inset
-
-.
- Using Equation 1.2.5.8:
-\end_layout
-
-\begin_layout Standard
-\begin_inset Formula 
-\begin{align*}
-52! & =2^{49}3^{23}5^{12}7^{8}11^{4}13^{4}17^{3}19^{2}23^{2}\cdot29\cdot31\cdot37\cdot41\cdot43\cdot47,\\
-39! & =2^{35}3^{18}5^{8}7^{5}11^{3}13^{3}17^{2}19^{2}\cdot23\cdot29\cdot31\cdot37\\
-13! & =2^{10}3^{5}5^{2}\cdot7\cdot11\cdot13\\
-\hline \binom{52}{13} & =2^{4}5^{2}7^{2}\cdot17\cdot23\cdot41\cdot43\cdot47.
-\end{align*}
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-\backslash
-rexerc5[05]
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-
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-Use Pascal's triangle to explain the fact that 
-\begin_inset Formula $11^{4}=14641$
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-
-\begin_inset Formula $11^{4}=(10+1)^{4}=\binom{4}{0}10^{4}1^{0}+\binom{4}{1}10^{3}1^{1}+\binom{4}{2}10^{2}1^{2}+\binom{4}{3}10^{1}1^{3}+\binom{4}{4}10^{0}1^{4}=14641$
-\end_inset
-
-.
- The pattern works for the number 11 in any given base until the number 10 or higher appears,
- which in this case happens in the next row.
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-
-\begin_layout Standard
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-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc6[10]
-\end_layout
-
-\end_inset
-
-Pascal's triangle (Table 1) can be extended in all directions by use of the addition formula,
- Eq.
- (9).
- Find the three rows that go on 
-\emph on
-top
-\emph default
- of Table 1 (i.e.
- for 
-\begin_inset Formula $r=-1$
-\end_inset
-
-,
- 
-\begin_inset Formula $-2$
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-,
- and 
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-The table is based on the properties that 
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- and that 
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- for 
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-.
- This implies that 
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-
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- (really for any 
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-) and that 
-\begin_inset Formula $\binom{r-1}{k}=\binom{r}{k}-\binom{r-1}{k-1}$
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-
- (the one below in the table minus the one on the left).
-\end_layout
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-\begin_layout Standard
-\align center
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-<lyxtabular version="3" rows="4" columns="11">
-<features tabularvalignment="middle">
-<column alignment="center" valignment="top">
-<column alignment="center" valignment="top">
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-<column alignment="center" valignment="top">
-<column alignment="center" valignment="top">
-<column alignment="center" valignment="top">
-<column alignment="center" valignment="top">
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-
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-
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-
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-
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-\begin_inset Formula $\binom{r}{6}$
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-
-
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-
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-
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-
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-
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-1
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-
-\begin_layout Plain Layout
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-
-
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-
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-
-\begin_layout Plain Layout
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-
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-
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-
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-
-
-\end_layout
-
-\end_inset
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-\begin_inset Text
-
-\begin_layout Plain Layout
-\begin_inset Formula $15$
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-
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-
-\end_inset
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-\begin_inset Text
-
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-\begin_inset Formula $-21$
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-
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-
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-\begin_inset Text
-
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-\begin_inset Formula $28$
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-
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-\begin_inset Text
-
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-\begin_inset Formula $-36$
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-
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-\begin_inset Text
-
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-\begin_inset Formula $45$
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-
-
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-\begin_inset Text
-
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-\begin_inset Formula $-55$
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-
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-
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-
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-1
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-\begin_inset Formula $-2$
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-
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-\begin_inset Text
-
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-\begin_inset Formula $3$
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-
-
-\end_layout
-
-\end_inset
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-
-
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-\begin_inset Text
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-
-
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-\begin_inset Text
-
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-\begin_inset Formula $7$
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-
-
-\end_layout
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-\end_inset
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-\begin_inset Text
-
-\begin_layout Plain Layout
-\begin_inset Formula $-8$
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-
-\end_layout
-
-\end_inset
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-\begin_inset Text
-
-\begin_layout Plain Layout
-\begin_inset Formula $9$
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-
-
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-
-\end_inset
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-\backslash
-exerc8[00]
-\end_layout
-
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-What property of Pascal's triangle is reflected in the 
-\begin_inset Quotes eld
-\end_inset
-
-symmetry condition,
-\begin_inset Quotes erd
-\end_inset
-
- Eq.
- (6)?
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-\end_layout
-
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-
-The symmetry of each row 
-\begin_inset Formula $r$
-\end_inset
-
- with center in 
-\begin_inset Formula $k=\frac{r}{2}$
-\end_inset
-
-.
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-\begin_inset ERT
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-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc9[01]
-\end_layout
-
-\end_inset
-
-What is the value of 
-\begin_inset Formula $\binom{n}{n}$
-\end_inset
-
-?
- (Consider all integers 
-\begin_inset Formula $n$
-\end_inset
-
-.)
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-For 
-\begin_inset Formula $n\geq0$
-\end_inset
-
-,
- 
-\begin_inset Formula $\binom{n}{n}=\frac{n!}{n!1!}=1$
-\end_inset
-
-.
- For 
-\begin_inset Formula $n<0$
-\end_inset
-
-,
- by definition 
-\begin_inset Formula $\binom{n}{n}=0$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc10[M25]
-\end_layout
-
-\end_inset
-
-If 
-\begin_inset Formula $p$
-\end_inset
-
- is prime,
- show that:
-\end_layout
-
-\begin_layout Enumerate
-\begin_inset Formula $\binom{n}{p}\equiv\left\lfloor \frac{n}{p}\right\rfloor \pmod p$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Enumerate
-\begin_inset Formula $\binom{p}{k}\equiv0\pmod p$
-\end_inset
-
-,
- for 
-\begin_inset Formula $1\leq k\leq p-1$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Enumerate
-\begin_inset Formula $\binom{p-1}{k}\equiv(-1)^{k}\pmod p$
-\end_inset
-
-,
- for 
-\begin_inset Formula $0\leq k\leq p-1$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Enumerate
-\begin_inset Formula $\binom{p+1}{k}\equiv0\pmod p$
-\end_inset
-
-,
- for 
-\begin_inset Formula $2\leq k\leq p-1$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Enumerate
-(É.
- Lucas,
- 1877.)
-\begin_inset Formula 
-\[
-\binom{n}{k}\equiv\binom{\lfloor n/p\rfloor}{\lfloor k/p\rfloor}\binom{n\bmod p}{k\bmod p}\pmod p.
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Enumerate
-If the representations of 
-\begin_inset Formula $n$
-\end_inset
-
- and 
-\begin_inset Formula $k$
-\end_inset
-
- in the 
-\begin_inset Formula $p$
-\end_inset
-
--ary number system are
-\begin_inset Formula 
-\begin{eqnarray*}
-\begin{array}{rlc}
-n & = & a_{r}p^{r}+\dots+a_{1}p+a_{0},\\
-k & = & b_{r}p^{r}+\dots+b_{1}p+b_{0},
-\end{array} & \text{then} & \binom{n}{k}\equiv\binom{a_{r}}{b_{r}}\cdots\binom{a_{1}}{b_{1}}\binom{a_{0}}{b_{0}}\pmod p.
-\end{eqnarray*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-We assume 
-\begin_inset Formula $n\in\mathbb{Z}$
-\end_inset
-
- and 
-\begin_inset Formula $k\in\mathbb{N}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Enumerate
-Using part 5,
-\begin_inset Formula 
-\[
-\binom{n}{p}\equiv\binom{\lfloor n/p\rfloor}{\lfloor p/p\rfloor}\binom{n\bmod p}{p\bmod p}=\binom{\lfloor n/p\rfloor}{1}\binom{n\bmod p}{0}=\left\lfloor \frac{n}{p}\right\rfloor \cdot1.
-\]
-
-\end_inset
-
-
-\begin_inset Note Note
-status open
-
-\begin_layout Plain Layout
-\begin_inset Note Comment
-status open
-
-\begin_layout Plain Layout
-Since we know 
-\begin_inset Formula $\binom{n}{p}=\frac{n(n-1)\cdots(n-p+1)}{1\cdot2\cdot\cdots\cdot p}\in\mathbb{Z}$
-\end_inset
-
- (because of the addition formula),
- and since 
-\begin_inset Formula $p$
-\end_inset
-
- appears once in the denominator,
- there must be a 
-\begin_inset Formula $k\in\{0,\dots,p-1\}$
-\end_inset
-
- such that 
-\begin_inset Formula $p\mid n-k$
-\end_inset
-
-,
- and it obviously must be unique.
- It's then easy to see that 
-\begin_inset Formula $\lfloor\frac{n}{p}\rfloor=\frac{n-k}{p}$
-\end_inset
-
-,
- and so
-\begin_inset Formula 
-\[
-\binom{n}{p}\equiv\left\lfloor \frac{n}{p}\right\rfloor \frac{n(n-1)\cdots(n-k+1)(n-k-1)\cdots(n-p+1)}{1\cdot2\cdots(p-1)}\pmod p.
-\]
-
-\end_inset
-
-Let us call 
-\begin_inset Formula $b$
-\end_inset
-
- the numerator in the fraction of the above formula,
- we just need to prove that 
-\begin_inset Formula $\frac{b}{(p-1)!}\equiv1\pmod p$
-\end_inset
-
-,
- which by rule 1.2.4–C is equivalent to 
-\begin_inset Formula $b\equiv(p-1)!\bmod p!$
-\end_inset
-
-.
- Since 
-\begin_inset Formula $b$
-\end_inset
-
- is a multiple of 
-\begin_inset Formula $(p-1)!$
-\end_inset
-
- (otherwise 
-\begin_inset Formula $\binom{n}{p}$
-\end_inset
-
- would be an integer),
- 
-\begin_inset Formula $b\equiv0\equiv(p-1)!\bmod(p-1)!$
-\end_inset
-
-,
- and since 
-\begin_inset Formula $\{(n-j)\bmod p\}_{0\leq j<p}^{j\neq k}=\{1,\dots,p-1\}$
-\end_inset
-
-,
- we have
-\begin_inset Formula 
-\[
-b=\prod_{\begin{subarray}{c}
-0\leq j<p\\
-j\neq k
-\end{subarray}}(n-j)\equiv\prod_{j=1}^{p-1}j=(p-1)!\pmod p.
-\]
-
-\end_inset
-
-Therefore,
- since 
-\begin_inset Formula $(p-1)!\bot p$
-\end_inset
-
-,
- rule 1.2.4–D gives us the result.
-\end_layout
-
-\end_inset
-
-
-\end_layout
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Enumerate
-\begin_inset Formula $\binom{p}{k}=\frac{p!}{k!(p-k)!}$
-\end_inset
-
-,
- but neither 
-\begin_inset Formula $k!$
-\end_inset
-
- nor 
-\begin_inset Formula $(p-k)!$
-\end_inset
-
- divide 
-\begin_inset Formula $p$
-\end_inset
-
- while 
-\begin_inset Formula $p!$
-\end_inset
-
- does,
- so 
-\begin_inset Formula $p\mid\binom{p}{k}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Enumerate
-We prove it by induction on 
-\begin_inset Formula $k$
-\end_inset
-
-.
- For 
-\begin_inset Formula $k=0$
-\end_inset
-
-,
- 
-\begin_inset Formula $\binom{p-1}{0}=1=(-1)^{0}$
-\end_inset
-
-,
- so the equation holds.
- For 
-\begin_inset Formula $1\leq k\leq p-1$
-\end_inset
-
-,
- 
-\begin_inset Formula 
-\[
-\binom{p-1}{k}=\binom{p}{k}-\binom{p-1}{k-1}\equiv0-(-1)^{k-1}=(-1)^{k}\pmod p,
-\]
-
-\end_inset
-
- where for the equivalence we use the previous part of the exercise and the induction hypothesis.
-\end_layout
-
-\begin_layout Enumerate
-Using part 2,
-\begin_inset Formula 
-\[
-\binom{p+1}{k}=\binom{p}{k}+\binom{p}{k-1}\equiv0-0=0\pmod k.
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Enumerate
-If 
-\begin_inset Formula $k<0$
-\end_inset
-
-,
- then 
-\begin_inset Formula $\binom{n}{k}=0$
-\end_inset
-
- and 
-\begin_inset Formula $\binom{\lfloor n/p\rfloor}{\lfloor k/p\rfloor}=0$
-\end_inset
-
-.
- Now let's prove this for 
-\begin_inset Formula $k\geq0$
-\end_inset
-
-.
- First,
- note that,
- if 
-\begin_inset Formula $a,b,c,d\in\mathbb{Z}$
-\end_inset
-
-,
- 
-\begin_inset Formula $\frac{a}{b},\frac{c}{d}\in\mathbb{Z}$
-\end_inset
-
-,
- and 
-\begin_inset Formula $a\equiv c$
-\end_inset
-
- and 
-\begin_inset Formula $0\not\equiv b\equiv d\pmod p$
-\end_inset
-
-,
- then 
-\begin_inset Formula 
-\[
-\frac{a}{b}\equiv\frac{c}{d}\pmod p.
-\]
-
-\end_inset
-
-Effectively,
- let 
-\begin_inset Formula $a=jp+b$
-\end_inset
-
- 
-\begin_inset Formula $c=kp+d$
-\end_inset
-
- for some integers 
-\begin_inset Formula $j$
-\end_inset
-
- and 
-\begin_inset Formula $k$
-\end_inset
-
-,
- then 
-\begin_inset Formula 
-\[
-\frac{a}{b}-\frac{c}{d}=\frac{ad-bc}{bd}=\frac{(jd-bk)p}{bd},
-\]
-
-\end_inset
-
-which is a multiple of 
-\begin_inset Formula $p$
-\end_inset
-
- because it's an integer and,
- since 
-\begin_inset Formula $bd\nmid p$
-\end_inset
-
-,
- it follows that 
-\begin_inset Formula $bd\mid jd-bk$
-\end_inset
-
-.
- Now,
- let 
-\begin_inset Formula $n\eqqcolon n'p+n_{0}$
-\end_inset
-
- and 
-\begin_inset Formula $k\eqqcolon k'p+k_{0}$
-\end_inset
-
- with 
-\begin_inset Formula $n',n_{0},k',k_{0}\in\mathbb{Z}$
-\end_inset
-
- and 
-\begin_inset Formula $0\leq n_{0},k_{0}<p$
-\end_inset
-
-.
- Clearly 
-\begin_inset Formula $n_{0}\equiv n$
-\end_inset
-
- and 
-\begin_inset Formula $k_{0}\equiv k$
-\end_inset
-
- (modulo 
-\begin_inset Formula $p$
-\end_inset
-
-),
- and then
-\begin_inset Formula 
-\begin{align*}
-\binom{n}{k} & =\prod_{j=1}^{k}\frac{n+1-j}{j}=\left(\prod_{j=1}^{pk'}\frac{n+1-j}{j}\right)\left(\prod_{j=1}^{k_{0}}\frac{pn'+n_{0}+1-pk'-j}{j+pk'}\right)\\
- & \equiv\left(\prod_{j=1}^{pk'}\frac{n+1-j}{j}\right)\left(\prod_{j=1}^{k_{0}}\frac{n_{0}+1-j}{j}\right)=\binom{n}{pk'}\binom{n_{0}}{k_{0}}\pmod p,
-\end{align*}
-
-\end_inset
-
-where for the equivalence we used that both 
-\begin_inset Formula $\binom{n}{k}$
-\end_inset
-
- and 
-\begin_inset Formula $\binom{n}{pk'}\binom{n_{0}}{k_{0}}$
-\end_inset
-
- are integers.
- Now we have to prove that 
-\begin_inset Formula $\binom{n}{pk'}\equiv\binom{n'}{k'}$
-\end_inset
-
-.
- This coefficient can be factored as
-\begin_inset Formula 
-\[
-\binom{n}{pk'}=\prod_{j=0}^{k'-1}\left(\prod_{i=1}^{p}\frac{n+1-jp-i}{jp+i}\right)=\prod_{j=0}^{k'-1}\prod_{i=1}^{p}\frac{(n'-j)p+n_{0}+1-i}{jp+i}.
-\]
-
-\end_inset
-
-Each of the 
-\begin_inset Formula $k'$
-\end_inset
-
- groups of factors has exactly one factor whose numerator is a multiple of 
-\begin_inset Formula $p$
-\end_inset
-
- (when 
-\begin_inset Formula $i=s\coloneqq(n_{0}+1)\bmod p$
-\end_inset
-
-),
- and one whose denominator is.
- Canceling the 
-\begin_inset Formula $p$
-\end_inset
-
- in them and taking congruences on the other factors,
- we get
-\begin_inset Formula 
-\begin{align*}
-\binom{n}{pk'} & \equiv\prod_{j=0}^{k'-1}\left(\frac{n'-j}{j+1}\frac{\prod_{\begin{subarray}{c}
-1\leq i\leq p\\
-i\neq s
-\end{subarray}}(n_{0}+1-i)}{(p-1)!}\right)=\prod_{j=0}^{k'-1}\left(\frac{n'-j}{j+1}\frac{(p-1)!}{(p-1)!}\right)=\\
- & =\prod_{j=1}^{k'}\frac{n'+1+j}{j}=\binom{n'}{k'}\pmod p.
-\end{align*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Enumerate
-From part 5 by induction on 
-\begin_inset Formula $r$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc11[M20]
-\end_layout
-
-\end_inset
-
-(E.
- Kummer,
- 1852.) Let 
-\begin_inset Formula $p$
-\end_inset
-
- be prime.
- Show that if 
-\begin_inset Formula $p^{n}$
-\end_inset
-
- divides
-\begin_inset Formula 
-\[
-\binom{a+b}{a}
-\]
-
-\end_inset
-
-but 
-\begin_inset Formula $p^{n+1}$
-\end_inset
-
- does not,
- then 
-\begin_inset Formula $n$
-\end_inset
-
- is equal to the number of 
-\emph on
-carries
-\emph default
- that occur when 
-\begin_inset Formula $a$
-\end_inset
-
- is added to 
-\begin_inset Formula $b$
-\end_inset
-
- in the 
-\begin_inset Formula $p$
-\end_inset
-
--ary number system.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Because of the way the exercise is written,
- we may assume that both 
-\begin_inset Formula $a$
-\end_inset
-
- and 
-\begin_inset Formula $b$
-\end_inset
-
- are natural numbers (non-negative integers),
- and therefore 
-\begin_inset Formula $\binom{a+b}{a}=\frac{(a+b)!}{a!b!}$
-\end_inset
-
-.
- Let 
-\begin_inset Formula $s_{a}$
-\end_inset
-
- be the sum of the digits of 
-\begin_inset Formula $a$
-\end_inset
-
- in base 
-\begin_inset Formula $p$
-\end_inset
-
- and let 
-\begin_inset Formula $\mu_{a}=\frac{a-s_{a}}{p-1}$
-\end_inset
-
- be the number of times 
-\begin_inset Formula $p$
-\end_inset
-
- appears in the prime factorization of 
-\begin_inset Formula $a$
-\end_inset
-
-,
- by Exercise 1.2.5–12.
- Define 
-\begin_inset Formula $s_{b}$
-\end_inset
-
-,
- 
-\begin_inset Formula $\mu_{b}$
-\end_inset
-
-,
- 
-\begin_inset Formula $s_{c}$
-\end_inset
-
-,
- and 
-\begin_inset Formula $\mu_{c}$
-\end_inset
-
- in a similar manner,
- with 
-\begin_inset Formula $c\coloneqq a+b$
-\end_inset
-
-.
- Then the number of times 
-\begin_inset Formula $p$
-\end_inset
-
- appears in the prime factorization of 
-\begin_inset Formula $\binom{a+b}{a}$
-\end_inset
-
- is precisely
-\begin_inset Formula 
-\[
-\mu_{c}-\mu_{a}-\mu_{b}=\frac{\cancel{c-a-b}-s_{c}+s_{a}+s_{b}}{p-1},
-\]
-
-\end_inset
-
-but 
-\begin_inset Formula $s_{c}$
-\end_inset
-
- is precisely the sum of 
-\begin_inset Formula $s_{a}$
-\end_inset
-
- plus 
-\begin_inset Formula $s_{b}$
-\end_inset
-
- except that,
- every time we do a carry in the sum in base 
-\begin_inset Formula $p$
-\end_inset
-
-,
- we have to subtract 
-\begin_inset Formula $p$
-\end_inset
-
- and add 1,
- that is,
- we subtract 
-\begin_inset Formula $p-1$
-\end_inset
-
-,
- so 
-\begin_inset Formula $s_{a}+s_{b}-s_{c}$
-\end_inset
-
- is precisely 
-\begin_inset Formula $p-1$
-\end_inset
-
- times the number of carries and this proves the result.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc17[M18]
-\end_layout
-
-\end_inset
-
-Prove the Chu-Vandermonde formula (21) from Eq.
- (15),
- using that 
-\begin_inset Formula $(1+x)^{r+s}=(1+x)^{r}(1+x)^{s}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-If 
-\begin_inset Formula $n$
-\end_inset
-
- is negative,
- either 
-\begin_inset Formula $k$
-\end_inset
-
- or 
-\begin_inset Formula $n-k$
-\end_inset
-
- is negative for any 
-\begin_inset Formula $k$
-\end_inset
-
- so both sides of the equation are 0.
- Otherwise,
- for 
-\begin_inset Formula $x\in(-1,1)$
-\end_inset
-
-,
-\begin_inset Formula 
-\begin{multline*}
-\sum_{n}\binom{r+s}{n}x^{n}=(1+x)^{r+s}=(1+x)^{r}(1+x)^{s}=\sum_{k}\binom{r}{k}x^{k}\sum_{m}\binom{s}{m}x^{m}=\\
-=\sum_{k}\binom{r}{k}x^{k}\sum_{n}\binom{s}{n-k}x^{n-k}=\sum_{n}\left(\sum_{k}\binom{r}{k}\binom{s}{n-k}\right)x^{n}.
-\end{multline*}
-
-\end_inset
-
-If 
-\begin_inset Formula $r$
-\end_inset
-
- and 
-\begin_inset Formula $s$
-\end_inset
-
- are integers,
- the sums are finite and therefore we have polynomials in 
-\begin_inset Formula $x$
-\end_inset
-
- in both side,
- but since they are both equal in a nonempty open interval,
- the coefficients are all equal and we have proved the result.
- If 
-\begin_inset Formula $r$
-\end_inset
-
- and 
-\begin_inset Formula $s$
-\end_inset
-
- are not integers,
- since
-\begin_inset Formula 
-\[
-\binom{r+s}{n}-\sum_{k}\binom{r}{k}\binom{s}{n-k}=0
-\]
-
-\end_inset
-
-for all 
-\begin_inset Formula $r,s\in\mathbb{Z}$
-\end_inset
-
-,
- and the left hand side is a polynomial on 
-\begin_inset Formula $r$
-\end_inset
-
- and 
-\begin_inset Formula $s$
-\end_inset
-
-,
- it follows that the polynomial is 0 everywhere.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc21[05]
-\end_layout
-
-\end_inset
-
-Both sides of Eq.
- (25) are polynomials in 
-\begin_inset Formula $s$
-\end_inset
-
-;
- why isn't that equation an identity in 
-\begin_inset Formula $s$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-We know the equation to be valid for 
-\begin_inset Formula $n+1$
-\end_inset
-
- values of 
-\begin_inset Formula $s$
-\end_inset
-
-,
- namely 
-\begin_inset Formula $0,1,\dots,n$
-\end_inset
-
-,
- but while the polynomial on the left has degree 
-\begin_inset Formula $n$
-\end_inset
-
-,
- the one on the right has degree 
-\begin_inset Formula $m+n+1\geq n+1$
-\end_inset
-
-,
- so we cannot apply the reasoning below Eq.
- (8).
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc30[M24]
-\end_layout
-
-\end_inset
-
-Show that there is a better way to solve Example 3 than the way used in the text,
- by manipulating the sum so that Eq.
- (26) applies.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-First,
- we note that the terms are nonzero when 
-\begin_inset Formula $k\geq0$
-\end_inset
-
- and 
-\begin_inset Formula $m+2k\leq n+k$
-\end_inset
-
-,
- which happens precisely when 
-\begin_inset Formula $0\leq k\leq n-m$
-\end_inset
-
-,
- and in particular we may assume 
-\begin_inset Formula $n\geq m$
-\end_inset
-
- since otherwise the sum is 0.
-\end_layout
-
-\begin_layout Standard
-We start by negating the upper index (rule G) in the second factor and using symmetry (rule B) on the first,
- and noting that all factors with negative 
-\begin_inset Formula $k$
-\end_inset
-
- are 0:
-\begin_inset Formula 
-\[
-\sum_{k}\binom{n+k}{m+2k}\binom{2k}{k}\frac{(-1)^{k}}{k+1}=\sum_{k\geq0}\binom{n+k}{n-m-k}\binom{-k-1}{k}\frac{1}{k+1}.
-\]
-
-\end_inset
-
-We have the left hand side of Eq.
- (26) with 
-\begin_inset Formula $t\mapsto1$
-\end_inset
-
-,
- 
-\begin_inset Formula $r\mapsto-1$
-\end_inset
-
-,
- 
-\begin_inset Formula $n\mapsto n-m$
-\end_inset
-
-,
- and 
-\begin_inset Formula $s\mapsto2n-m$
-\end_inset
-
-,
- so this is equal to
-\begin_inset Formula 
-\[
-\binom{-1+2n-m-n+m}{n-m}=\binom{n-1}{n-m}=\binom{n-1}{m-1}.
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc31[M20]
-\end_layout
-
-\end_inset
-
-Evaluate
-\begin_inset Formula 
-\[
-\sum_{k}\binom{m-r+s}{k}\binom{n+r-s}{n-k}\binom{r+k}{m+n}
-\]
-
-\end_inset
-
-in terms of 
-\begin_inset Formula $r$
-\end_inset
-
-,
- 
-\begin_inset Formula $s$
-\end_inset
-
-,
- 
-\begin_inset Formula $m$
-\end_inset
-
-,
- and 
-\begin_inset Formula $n$
-\end_inset
-
-,
- given that 
-\begin_inset Formula $m$
-\end_inset
-
- and 
-\begin_inset Formula $n$
-\end_inset
-
- are integers.
- Begin by replacing
-\begin_inset Formula 
-\begin{eqnarray*}
-\binom{r+k}{m+n} & \text{ by } & \sum_{j}\binom{r}{m+n-j}\binom{k}{j}.
-\end{eqnarray*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-We start assuming that 
-\begin_inset Formula $r$
-\end_inset
-
- and 
-\begin_inset Formula $s$
-\end_inset
-
- are also integers.
- The replacement suggested is given directly by Equation 21,
- and we get
-\begin_inset Formula 
-\begin{align*}
-S & \coloneqq\sum_{k}\binom{m-r+s}{k}\binom{n+r-s}{n-k}\binom{r+k}{m+n}\\
- & =\sum_{k}\sum_{j}\binom{m-r+s}{k}\binom{n+r-s}{n-k}\binom{r}{m+n-j}\binom{k}{j} & \text{Eq. (21)}\\
- & =\sum_{j}\binom{r}{m+n-j}\binom{m-r+s}{j}\sum_{k}\binom{n+r-s}{n-k}\binom{m-r+s-j}{k-j} & \text{Eq. (20)}\\
- & =\sum_{j}\binom{r}{m+n-j}\binom{m-r+s}{j}\sum_{k}\binom{n+r-s}{k+r-s}\binom{m-r+s-j}{k-j} & \text{Eq. (6)*}\\
- & =\sum_{j}\binom{r}{m+n-j}\binom{m-r+s}{j}\binom{m+n-j}{n-j} & \text{Eq. (22)*}\\
- & =\sum_{j}\binom{r}{m+n-j}\binom{m-r+s}{j}\binom{m+n-j}{m} & \text{Eq. (6)**}\\
- & =\binom{r}{m}\sum_{j}\binom{m-r+s}{j}\binom{r-m}{n-j} & \text{Eq. (20)}\\
- & =\binom{r}{m}\binom{s}{n}.
-\end{align*}
-
-\end_inset
-
- In (**),
- we use that,
- for nonzero addends,
- 
-\begin_inset Formula $m+n-j\geq0$
-\end_inset
-
- and therefore we can apply Equation 6.
- In (*),
- we assume 
-\begin_inset Formula $s\leq m-r$
-\end_inset
-
-.
- This is not important,
- since,
- by the reasoning about polynomials,
- this has been proved for infinite values of 
-\begin_inset Formula $r$
-\end_inset
-
- and infinite values of 
-\begin_inset Formula $s$
-\end_inset
-
- and therefore applies to all 
-\begin_inset Formula $r$
-\end_inset
-
- and 
-\begin_inset Formula $s$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc36[M10]
-\end_layout
-
-\end_inset
-
-What is the sum 
-\begin_inset Formula $\sum_{k}\binom{n}{k}$
-\end_inset
-
- of the numbers in each row of Pascal's triangle?
- What is the sum of these numbers with alternating signs,
- 
-\begin_inset Formula $\sum_{k}\binom{n}{k}(-1)^{k}$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-If 
-\begin_inset Formula $n\geq0$
-\end_inset
-
-,
-\begin_inset Formula 
-\begin{align*}
-\sum_{k}\binom{n}{k}=\sum_{k}\binom{n}{k}1^{k}1^{n-k} & =(1+1)^{n}=2^{n},\\
-\sum_{k}\binom{n}{k}(-1)^{k}=\sum_{k}\binom{n}{k}(-1)^{k}1^{n-k} & =(1-1)^{n}=0^{n}=\delta_{n0}.
-\end{align*}
-
-\end_inset
-
-If 
-\begin_inset Formula $n<0$
-\end_inset
-
- the value is generally undefined,
- as evidenced by Exercise 6.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc37[M10]
-\end_layout
-
-\end_inset
-
-From the answers to the preceding exercise,
- deduce the value of the sum of every other entry in a row,
- 
-\begin_inset Formula $\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\dots$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer
-\end_layout
-
-\end_inset
-
-
-\begin_inset Formula 
-\[
-\sum_{k}\binom{n}{2k}=\sum_{k}\binom{n}{k}[k\text{ even}]=\sum_{k}\binom{n}{k}\frac{1+(-1)^{k}}{2}=\frac{2^{n}+\delta_{n0}}{2}=\left\{ \begin{array}{rl}
-1, & n=0;\\
-2^{n-1}, & n>0.
-\end{array}\right.
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc39[M10]
-\end_layout
-
-\end_inset
-
-What is the sum 
-\begin_inset Formula $\sum_{k}\stirla nk$
-\end_inset
-
- of the numbers in each row of Stirling's first triangle?
- What is the sum of these numbers with alternating signs?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-We can use Equation 44:
-\end_layout
-
-\begin_layout Standard
-\begin_inset Formula 
-\begin{multline*}
-\sum_{k}\stirla nk=(-1)^{-n}\sum_{k}(-1)^{n-k}\stirla nk(-1)^{k}=(-1)^{n}(-1)^{\underline{n}}=\\
-=(-1)^{n}(-1)(-2)\cdots(-n)=n!,
-\end{multline*}
-
-\end_inset
-
-
-\begin_inset Formula 
-\[
-\sum_{k}\stirla nk(-1)^{k}=\sum_{k}\stirla nk(-1)^{-k}=(-1)^{-n}\sum_{k}(-1)^{n-k}\stirla nk1^{k}=(-1)^{n}1^{\underline{n}}=\delta_{n0}-\delta_{n1}.
-\]
-
-\end_inset
-
-The first equation makes sense,
- as it is the number of permutations of 
-\begin_inset Formula $n$
-\end_inset
-
- symbols irrespective of the number 
-\begin_inset Formula $k$
-\end_inset
-
- of cycles.
- 
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc42[HM10]
-\end_layout
-
-\end_inset
-
-Express the binomial coefficient 
-\begin_inset Formula $\binom{r}{k}$
-\end_inset
-
- in terms of the beta function defined above.
- (This gives us a way to extend the definition to all real values of 
-\begin_inset Formula $k$
-\end_inset
-
-.)
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-For integers 
-\begin_inset Formula $r\geq0$
-\end_inset
-
- and 
-\begin_inset Formula $k\neq0$
-\end_inset
-
-,
- 
-\begin_inset Formula 
-\[
-\binom{r}{k}=\frac{r!}{k!(r-k)!}=\frac{\Gamma(r+1)}{\Gamma(k+1)\Gamma(r-k)}=\frac{\Gamma(r+1)}{k\Gamma(k)\Gamma(r-k+1)}=(kB(k,r-k+1))^{-1}.
-\]
-
-\end_inset
-
-We can extend this function to the points where it is not defined by taking limits.
-\end_layout
-
-\begin_layout Standard
-We are not required to evaluate the domain of this function,
- but we have to see that it matches our definition for 
-\begin_inset Formula $r\in\mathbb{R}$
-\end_inset
-
- and 
-\begin_inset Formula $k\in\mathbb{Z}$
-\end_inset
-
-.
- For 
-\begin_inset Formula $k\in\mathbb{N}$
-\end_inset
-
-,
- we first see that 
-\begin_inset Formula 
-\begin{multline*}
-\lim_{(x,y)\to(r,k)}\frac{\Gamma(x-y+1)}{\Gamma(x-k+1)}=\lim_{(x,y)\to(r,k)}\frac{\lim_{m}\frac{m^{x-y}m!}{(x-y+1)\cdots(x-y+m)}}{\lim_{m}\frac{m^{x-k}m!}{(x-k+1)\cdots(x-k+m)}}=\\
-=\lim_{(x,y)\to(r,k)}\lim_{m}m^{k-y}\prod_{i=1}^{m}\frac{x+m-y}{x+m-k}=1,
-\end{multline*}
-
-\end_inset
-
-where I don't remember enough of my calculus lessons to know why the last step is correct but it has to do with exchanging limits.
- With this,
- if we successively apply the fact that 
-\begin_inset Formula $\Gamma(x+1)=x\Gamma(x)$
-\end_inset
-
-,
- 
-\begin_inset Formula 
-\begin{multline*}
-\lim_{(x,y)\to(r,k)}\frac{\Gamma(x+1)}{y\Gamma(y)\Gamma(x-y+1)}=\lim_{(x,y)\to(r,k)}\frac{x(x-1)\cdots(x-k+1)}{y\Gamma(y)\frac{\Gamma(x-y+1)}{\Gamma(x-k+1)}}=\\
-=\frac{r(r-1)\cdots(r-k+1)}{k!}.
-\end{multline*}
-
-\end_inset
-
-If 
-\begin_inset Formula $k\in\mathbb{Z}^{-}$
-\end_inset
-
-,
- for 
-\begin_inset Formula $r\notin\mathbb{Z}^{-}$
-\end_inset
-
-,
- 
-\begin_inset Formula $\Gamma(r+1)$
-\end_inset
-
- is bounded in an open set around 
-\begin_inset Formula $(r,k)$
-\end_inset
-
-,
- and since 
-\begin_inset Formula $\Gamma$
-\end_inset
-
- is continuous and has no zeros,
- the denominator tends to infinity,
- so the corresponding limit correctly evaluates to 0.
- If 
-\begin_inset Formula $k,r\in\mathbb{Z}^{-}$
-\end_inset
-
- I can't find if the value is defined,
- but if it is then its value is 0.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc44[HM20]
-\end_layout
-
-\end_inset
-
-Using the generalized binomial coefficient suggested in exercise 42,
- show that
-\begin_inset Formula 
-\[
-\binom{r}{1/2}=2^{2r+1}\bigg/\binom{2r}{r}\pi.
-\]
-
-\end_inset
-
-
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-From the definition,
-\begin_inset Formula 
-\[
-\binom{r}{\frac{1}{2}}=\frac{1}{\frac{1}{2}B(\frac{1}{2},r+\frac{1}{2})}=\frac{2\Gamma(r+1)}{\Gamma(\frac{1}{2})\Gamma(r+\frac{1}{2})}.
-\]
-
-\end_inset
-
-Now,
- we know 
-\begin_inset Formula $\frac{\sqrt{\pi}}{2}=(\frac{1}{2})!=\frac{1}{2}\Gamma(\frac{1}{2})$
-\end_inset
-
-,
- so 
-\begin_inset Formula $\Gamma(\frac{1}{2})=\sqrt{\pi}$
-\end_inset
-
- and
-\begin_inset Formula 
-\[
-\binom{r}{\frac{1}{2}}=\frac{2r!}{\sqrt{\pi}(r-\frac{1}{2})(r-\frac{3}{2})\cdots(\frac{1}{2})\sqrt{\pi}}=\frac{2r!2^{r}}{\pi(2r-1)(2r-3)\cdots1}.
-\]
-
-\end_inset
-
-We see that 
-\begin_inset Formula $(2r)!=((2r)(2r-2)\cdots2)((2r-1)(2r-3)\cdots1)$
-\end_inset
-
-.
- We already have the second factor,
- and the first one is precisely 
-\begin_inset Formula $2^{r}r!$
-\end_inset
-
-,
- so we multiply both numerator and denominator by this factor to get that
-\begin_inset Formula 
-\[
-\binom{r}{\frac{1}{2}}=\frac{2r!2^{r}2^{r}r!}{\pi(2r)!}=\frac{2^{2r+1}}{\binom{2r}{r}\pi}.
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc46[M21]
-\end_layout
-
-\end_inset
-
-Using Stirling's approximation,
- Eq.
- 1.2.5–(7),
- find an approximate value of 
-\begin_inset Formula $\binom{x+y}{y}$
-\end_inset
-
-,
- assuming that both 
-\begin_inset Formula $x$
-\end_inset
-
- and 
-\begin_inset Formula $y$
-\end_inset
-
- are large.
- In particular,
- find the approximate size of 
-\begin_inset Formula $\binom{2n}{n}$
-\end_inset
-
- when 
-\begin_inset Formula $n$
-\end_inset
-
- is large.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-For large 
-\begin_inset Formula $x$
-\end_inset
-
- and 
-\begin_inset Formula $y$
-\end_inset
-
-,
-\begin_inset Formula 
-\begin{multline*}
-\binom{x+y}{y}=\frac{(x+y)!}{x!y!}\approx\frac{\sqrt{2\pi(x+y)}\left(\frac{x+y}{\text{e}}\right)^{x+y}}{\sqrt{2\pi x}\sqrt{2\pi y}\left(\frac{x}{\text{e}}\right)^{x}\left(\frac{y}{\text{e}}\right)^{y}}=\sqrt{\frac{x+y}{2\pi xy}}\left(\frac{(x+y)^{x+y}}{x^{x}y^{y}}\right)=\\
-=\frac{(x+y)^{x+y+\nicefrac{1}{2}}}{\sqrt{2\pi}x^{x+\nicefrac{1}{2}}y^{y+\nicefrac{1}{2}}}=\sqrt{\frac{1}{2\pi}\left(\frac{1}{x}+\frac{1}{y}\right)}\left(1+\frac{y}{x}\right)^{x}\left(1+\frac{x}{y}\right)^{y}.
-\end{multline*}
-
-\end_inset
-
-Then,
- for large 
-\begin_inset Formula $n$
-\end_inset
-
-,
-\begin_inset Formula 
-\[
-\binom{2n}{n}=\frac{(2n)^{2n+\nicefrac{1}{2}}}{\sqrt{2\pi}n^{2n+1}}=\frac{2^{2n}\sqrt{2}n^{2n}\sqrt{n}}{\sqrt{2\pi}n^{2n}n}=\frac{4^{n}}{\sqrt{\pi n}}.
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc48[M25]
-\end_layout
-
-\end_inset
-
-Show that
-\begin_inset Formula 
-\[
-\sum_{k\geq0}\binom{n}{k}\frac{(-1)^{k}}{k+x}=\frac{n!}{x(x+1)\cdots(x+n)}=\frac{1}{x\binom{n+x}{n}},
-\]
-
-\end_inset
-
-if the denominators are not zero.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-The second equality is obvious.
- For the first one,
- since the way the formula is written assumes 
-\begin_inset Formula $n\in\mathbb{N}$
-\end_inset
-
-,
- we can prove this by induction.
- For 
-\begin_inset Formula $n=0$
-\end_inset
-
-,
-\begin_inset Formula 
-\[
-\sum_{k\geq0}\binom{0}{k}\frac{(-1)^{k}}{k+x}=\frac{1}{x}=\frac{0!}{x}.
-\]
-
-\end_inset
-
-For 
-\begin_inset Formula $n>0$
-\end_inset
-
-,
-\begin_inset Formula 
-\begin{multline*}
-\sum_{k\geq0}\binom{n}{k}\frac{(-1)^{k}}{k+x}=\sum_{k\geq0}\binom{n-1}{k}\frac{(-1)^{k}}{k+x}+\sum_{k\geq0}\binom{n-1}{k-1}\frac{(-1)^{k}}{k+x}=\\
-=\sum_{k\geq0}\binom{n-1}{k}\frac{(-1)^{k}}{k+x}+\sum_{k\geq0}\binom{n-1}{k}\frac{(-1)^{k+1}}{k+1+x}=\frac{1}{x\binom{n-1+x}{n-1}}-\frac{1}{(x+1)\binom{n+x}{n-1}}=\\
-=\frac{(n-1)!}{x(x+1)\cdots(x+n-1)}-\frac{(n-1)!}{(x+1)(x+2)\cdots(x+n)}=\frac{(n-1)!((\cancel{x+}n)\cancel{-x})}{x(x+1)\cdots(x+n)}.
-\end{multline*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc57[M22]
-\end_layout
-
-\end_inset
-
-Show that the coefficient 
-\begin_inset Formula $a_{m}$
-\end_inset
-
- in Stirling's attempt at generalizing the factorial function,
- Eq.
- 1.2.5–(12),
- is
-\begin_inset Formula 
-\[
-\frac{(-1)^{m}}{m!}\sum_{k\geq1}(-1)^{k}\binom{m-1}{k-1}\ln k.
-\]
-
-\end_inset
-
-Also find 
-\begin_inset Formula $q$
-\end_inset
-
--nomial generalizations of the fundamental identities (17) and (21).
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-The sum in Stirling's attempt is
-\begin_inset Formula 
-\begin{eqnarray*}
-S_{n} & = & \sum_{k\geq0}a_{k+1}\prod_{j=0}^{k}(n-j)=\sum_{m\geq1}a_{m}\prod_{j=0}^{m-1}(n-j)\\
- & = & \sum_{m\geq1}\frac{(-1)^{m}}{m!}\left(\sum_{k\geq1}(-1)^{k}\binom{m-1}{k-1}\ln k\right)\prod_{j=0}^{m-1}(n-j)\\
- & = & \sum_{k\geq1}\left(\sum_{m\geq1}(-1)^{m+k}\binom{n}{m}\binom{m-1}{k-1}\right)\ln k.
-\end{eqnarray*}
-
-\end_inset
-
-There are several things happening in the last line.
- First,
- note that both sums are bounded,
- for 
-\begin_inset Formula $m$
-\end_inset
-
- must be at most 
-\begin_inset Formula $n$
-\end_inset
-
- because otherwise the product would contain a factor 
-\begin_inset Formula $n-n=0$
-\end_inset
-
-,
- and 
-\begin_inset Formula $k$
-\end_inset
-
- must be lower than 
-\begin_inset Formula $m$
-\end_inset
-
-,
- and therefore lower than 
-\begin_inset Formula $n$
-\end_inset
-
-,
- because otherwise the factor 
-\begin_inset Formula $\binom{m-1}{k-1}$
-\end_inset
-
- would be 0.
- This allows us to swap the sums and extract a factor that only depends on 
-\begin_inset Formula $k$
-\end_inset
-
-.
- Moreover,
-\begin_inset Formula 
-\[
-\frac{\prod_{j=0}^{m-1}(n-j)}{m!}=\frac{\prod_{j=1}^{m}(n-j+1)}{\prod_{j=1}^{m}j}=\binom{n}{m}.
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-Let 
-\begin_inset Formula $A_{k}$
-\end_inset
-
- be the value of the inner sum for a given 
-\begin_inset Formula $k$
-\end_inset
-
-,
- we know that 
-\begin_inset Formula $A_{k}=0$
-\end_inset
-
- for 
-\begin_inset Formula $k>n$
-\end_inset
-
-,
- and if we prove that 
-\begin_inset Formula $A_{k}=1$
-\end_inset
-
- for 
-\begin_inset Formula $k\leq n$
-\end_inset
-
-,
- we would have
-\begin_inset Formula 
-\[
-S_{n}=\sum_{k=1}^{n}\ln k=\ln\prod_{k=1}^{n}k=\ln n!
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-Now,
- we have
-\begin_inset Formula 
-\begin{align*}
-A_{k} & =\sum_{m\geq1}(-1)^{m+k}\binom{n}{m}\binom{m-1}{k-1}\\
- & =\sum_{m}(-1)^{m+k}\binom{n}{m}\binom{m-1}{k-1}-(-1)^{k}\binom{-1}{k-1} & (*)\\
- & =\sum_{m}(-1)^{n-m-k}\binom{n}{n-m}\binom{n-m-1}{k-1}+1 & (**)\\
- & =-\sum_{m}(-1)^{n-m}\binom{n}{m}\binom{k-1-n+m}{k-1}+1 & \text{Rules B, G}\\
- & =-\binom{k-1-n}{k-1-n}+1=1, & \text{Eq. (23)}
-\end{align*}
-
-\end_inset
-
-where in 
-\begin_inset Formula $(*)$
-\end_inset
-
- we expand the domain of 
-\begin_inset Formula $m$
-\end_inset
-
- to all integers and subtract the case when 
-\begin_inset Formula $m=0$
-\end_inset
-
- to compensate for that (this is the only nonzero term that wasn't considered already) and in 
-\begin_inset Formula $(**)$
-\end_inset
-
- we change 
-\begin_inset Formula $m$
-\end_inset
-
- to 
-\begin_inset Formula $n-m$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-For the second part of the exercise,
- we start by writing the identities (17) and (21) in 
-\begin_inset Formula $q$
-\end_inset
-
--nomial notation.
- For (17),
- this would be
-\begin_inset Formula 
-\begin{eqnarray*}
-\binom{r}{k,r-k}=(-1)^{k}\binom{k-r-1}{k,-r-1}, & \text{ or } & \binom{a+b}{a,b}=(-1)^{a}\binom{-b-1}{a,-(a+b)-1}.
-\end{eqnarray*}
-
-\end_inset
-
-That is,
- this allows us to change the sum on top to something resembling just one of the numbers that were on the bottom part.
- To generalize this,
-\begin_inset Formula 
-\begin{align*}
-\binom{k_{1}+\dots+k_{m}}{k_{1},\dots,k_{m}} & =\binom{k_{1}+k_{2}}{k_{1}}\cdots\binom{k_{1}+\dots+k_{m}}{k_{1}+\dots+k_{m-1}}\\
- & =(-1)^{k_{1}+\dots+k_{m-1}}\binom{k_{1}+k_{2}}{k_{1}}\cdots\binom{k_{1}+\dots+k_{m-1}}{k_{1}+\dots+k_{m-2}}\binom{-k_{m}-1}{k_{1}+\dots+k_{m-1}}\\
- & =(-1)^{k_{1}+\dots+k_{m-1}}\binom{-k_{m}-1}{k_{1},\dots,k_{m-1},-(k_{1}+\dots+k_{m})-1}.
-\end{align*}
-
-\end_inset
-
-We are abusing notation here,
- since the book only defines multinomial coefficients for 
-\begin_inset Formula $k_{i}\geq0$
-\end_inset
-
-,
- but we can define them for any 
-\begin_inset Formula $k_{i}\in\mathbb{Z}$
-\end_inset
-
- by using the property that relates them to binomial coefficients,
- at the cost of symmetry with respect to the order of the 
-\begin_inset Formula $k_{i}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-For (21),
- this would be
-\begin_inset Formula 
-\[
-\sum_{k}\binom{r}{k,r-k}\binom{s}{n-k,s-n+k}=\binom{r+s}{n,r+s-n}.
-\]
-
-\end_inset
-
-That is,
- in this case we would add up all the numbers on each component.
- This clearly holds if we place 
-\begin_inset Formula $k,r-k$
-\end_inset
-
- and 
-\begin_inset Formula $n-k,s-n+k$
-\end_inset
-
- as the two last components of the multinomial coefficient.
- There must be a better (less silly) generalization but I've already spent way too much time in this exercise.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc60[M23]
-\end_layout
-
-\end_inset
-
-We have seen that 
-\begin_inset Formula $\binom{n}{k}$
-\end_inset
-
- is the number of combinations of 
-\begin_inset Formula $n$
-\end_inset
-
- things,
- 
-\begin_inset Formula $k$
-\end_inset
-
- at a time,
- namely the number of ways to choose 
-\begin_inset Formula $k$
-\end_inset
-
- different things out of a set of 
-\begin_inset Formula $n$
-\end_inset
-
-.
- The 
-\emph on
-combinations with repetitions
-\emph default
- are similar to ordinary combinations,
- except that we may choose each object any number of times.
- Thus,
- the list (1) would be extended to include also 
-\begin_inset Formula $aaa$
-\end_inset
-
-,
- 
-\begin_inset Formula $aab$
-\end_inset
-
-,
- 
-\begin_inset Formula $aac$
-\end_inset
-
-,
- 
-\begin_inset Formula $aad$
-\end_inset
-
-,
- 
-\begin_inset Formula $aae$
-\end_inset
-
-,
- 
-\begin_inset Formula $abb$
-\end_inset
-
-,
- etc.,
- if we were considering combinations with repetition.
- How many 
-\begin_inset Formula $k$
-\end_inset
-
--combinations of 
-\begin_inset Formula $n$
-\end_inset
-
- objects are there,
- if repetition is allowed?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-We can draw a bijection between 
-\begin_inset Formula $k$
-\end_inset
-
--combinations of 
-\begin_inset Formula $\{1,\dots,n\}$
-\end_inset
-
- with repetition and 
-\begin_inset Formula $k$
-\end_inset
-
--combinations of 
-\begin_inset Formula $\{1,\dots,n+k-1\}$
-\end_inset
-
- without repetition by the following assignment:
-\begin_inset Formula 
-\begin{eqnarray*}
-1\leq a_{1}\leq a_{2}\leq\dots\leq a_{k}\leq n & \mapsto & 1\leq a_{1}<a_{2}+1<\dots<a_{k}+k-1\leq n+k-1\\
-1\leq b_{1}\leq b_{2}-1\leq\dots\leq b_{k}-k+1\leq n & \mapsfrom & 1\leq b_{1}<b_{2}<\dots<b_{k}\leq n+k-1
-\end{eqnarray*}
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-Thus the answer is
-\begin_inset Formula 
-\[
-\binom{n+k-1}{k}=(-1)^{k}\binom{-n}{k}=\frac{n^{\overline{k}}}{k!}.
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc62[M23]
-\end_layout
-
-\end_inset
-
-The text gives formulas for sums involving a product of two binomial coefficients.
- Of the sums involving a product of three binomial coefficients,
- the following one and the identity of exercise 31 seem to be the most useful:
-\begin_inset Formula 
-\begin{align*}
-\sum_{k}(-1)^{k}\binom{l+m}{l+k}\binom{m+n}{m+k}\binom{n+l}{n+k} & =\frac{(l+m+n)!}{l!m!n!}, & \text{integer }l,m,n & \geq0.
-\end{align*}
-
-\end_inset
-
-(The sum includes both positive and negative values of 
-\begin_inset Formula $k$
-\end_inset
-
-.) Prove this identity.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
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-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-
-\begin_inset Note Greyedout
-status open
-
-\begin_layout Plain Layout
-(Looking at the solution.)
-\end_layout
-
-\end_inset
-
-We may assume that 
-\begin_inset Formula $l\leq m,n$
-\end_inset
-
-,
- since we can always rename coefficients to make it this way.
- Then the application of the identity in exercise 31 is actually from right to left:
- we add a new variable 
-\begin_inset Formula $j$
-\end_inset
-
-,
- change the last factor 
-\begin_inset Formula $\binom{n+l}{n+k}$
-\end_inset
-
- to 
-\begin_inset Formula $\binom{n+k}{l-k}$
-\end_inset
-
- by symmetry,
- and apply the aforementioned exercise to the last two factors to get that
-\begin_inset Formula 
-\[
-S\coloneqq\sum_{k}(-1)^{k}\binom{l+m}{l+k}\binom{m+n}{m+k}\binom{n+l}{n+k}=\sum_{j,k}(-1)^{k}\binom{l+m}{k+l}\binom{k+l}{j}\binom{m-k}{l-k-j}\binom{m+n+j}{m+l}.
-\]
-
-\end_inset
-
-Using that in general 
-\begin_inset Formula $\binom{n}{k}=0$
-\end_inset
-
- for 
-\begin_inset Formula $n\geq0$
-\end_inset
-
- and 
-\begin_inset Formula $k\notin\{0,\dots,n\}$
-\end_inset
-
-,
- nonzero addends here must follow
-\begin_inset Formula 
-\begin{align*}
--l & \leq k\leq m, & -m & \leq k\leq n, & -n & \leq k\leq l, & 0 & \leq j\leq l+k, & l-m & \leq j\leq l-k;
-\end{align*}
-
-\end_inset
-
-that is,
- 
-\begin_inset Formula $\vert k\vert\leq\min\{m,n,l\}$
-\end_inset
-
- and 
-\begin_inset Formula $\max\{l-m,0\}\leq j\leq l-\vert k\vert$
-\end_inset
-
-,
- which simplifies to 
-\begin_inset Formula $\vert k\vert\leq l$
-\end_inset
-
- and 
-\begin_inset Formula $0\leq j\leq l-\vert k\vert$
-\end_inset
-
- using the conditions at the beginning and then to only 
-\begin_inset Formula $0\leq j\leq l-\vert k\vert$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-This means that,
- in all the binomial coefficients in the right hand side of the above equation,
- the upper and lower parts are non-negative,
- so we can substitute them to fractions of coefficients and,
- after canceling out,
- we get
-\begin_inset Formula 
-\begin{align*}
-S & =\sum_{0\leq j\leq l-\vert k\vert}(-1)^{k}\frac{(m+n+j)!}{j!(l+k-j)!(l-k-j)!(m-l+j)!(n-l+j)!}\\
- & =\sum_{0\leq j\leq l}\sum_{\vert k\vert\leq l-j}(-1)^{k}\binom{2l-2j}{l+k-j}\frac{(m+n+j)!}{(2l-2j)!j!(m-l+j)!(n-l+j)!},
-\end{align*}
-
-\end_inset
-
-but then the sum of 
-\begin_inset Formula $(-1)^{k}\binom{2l-2j}{l+k-j}$
-\end_inset
-
- across all 
-\begin_inset Formula $k$
-\end_inset
-
- is 0 unless 
-\begin_inset Formula $j=l$
-\end_inset
-
-,
- so we just take the term with 
-\begin_inset Formula $j=l$
-\end_inset
-
- and 
-\begin_inset Formula $k=0$
-\end_inset
-
- to get
-\begin_inset Formula 
-\[
-S=(-1)^{0}\binom{2l-2l}{l+0-l}\frac{(m+n+l)!}{(2l-2l)!l!(m-l+l)!(n-l+l)!}=\frac{(l+m+n)!}{l!m!n!}.
-\]
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc64[M20]
-\end_layout
-
-\end_inset
-
-Show that 
-\begin_inset Formula $\stirlb nm$
-\end_inset
-
- is the number of ways to partition a set of 
-\begin_inset Formula $n$
-\end_inset
-
- elements into 
-\begin_inset Formula $m$
-\end_inset
-
- nonempty disjoint subsets.
- For example,
- the set 
-\begin_inset Formula $\{1,2,3,4\}$
-\end_inset
-
- can be partitioned into two subsets in 
-\begin_inset Formula $\stirlb 42=7$
-\end_inset
-
- ways:
- 
-\begin_inset Formula $\{1,2,3\}\{4\}$
-\end_inset
-
-;
- 
-\begin_inset Formula $\{1,2,4\}\{3\}$
-\end_inset
-
-;
- 
-\begin_inset Formula $\{1,3,4\}\{2\}$
-\end_inset
-
-;
- 
-\begin_inset Formula $\{2,3,4\}\{1\}$
-\end_inset
-
-;
- 
-\begin_inset Formula $\{1,2\}\{3,4\}$
-\end_inset
-
-;
- 
-\begin_inset Formula $\{1,3\}\{2,4\}$
-\end_inset
-
-;
- 
-\begin_inset Formula $\{1,4\}\{2,3\}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Let 
-\begin_inset Formula $n,m\in\mathbb{N}$
-\end_inset
-
-,
- we prove this by induction in 
-\begin_inset Formula $n$
-\end_inset
-
- and 
-\begin_inset Formula $m$
-\end_inset
-
-.
- For 
-\begin_inset Formula $n=0$
-\end_inset
-
-,
- there is 
-\begin_inset Formula $1=\stirlb 00$
-\end_inset
-
- way to partition 
-\begin_inset Formula $\emptyset$
-\end_inset
-
- into 0 nonempty disjoint subsets and 
-\begin_inset Formula $0=\stirlb 0m$
-\end_inset
-
- ways to partition it into 
-\begin_inset Formula $m\geq1$
-\end_inset
-
- nonempty subsets,
- which matches Equation 48.
- For 
-\begin_inset Formula $m=0$
-\end_inset
-
- and 
-\begin_inset Formula $n\geq1$
-\end_inset
-
-,
- there are 
-\begin_inset Formula $0=\stirlb n0$
-\end_inset
-
- ways to divide a set of 
-\begin_inset Formula $n$
-\end_inset
-
- elements into 0 nonempty subsets.
- 
-\end_layout
-
-\begin_layout Standard
-Now,
- for the induction step,
- if 
-\begin_inset Formula $n\geq1$
-\end_inset
-
- and 
-\begin_inset Formula $m\geq1$
-\end_inset
-
-,
- a partition of a set like 
-\begin_inset Formula $\{1,\dots,n\}$
-\end_inset
-
- into nonempty subsets can be thought of as a partition of 
-\begin_inset Formula $\{1,\dots,n-1\}$
-\end_inset
-
- to which 
-\begin_inset Formula $n$
-\end_inset
-
- has been added either to one of the elements of a partition or in a separate singleton set.
- If the partition of 
-\begin_inset Formula $\{1,\dots,n\}$
-\end_inset
-
- has 
-\begin_inset Formula $m$
-\end_inset
-
- nonempty subsets,
- for the first option we might add 
-\begin_inset Formula $n$
-\end_inset
-
- to any of the subsets of a partition of 
-\begin_inset Formula $\{1,\dots,n-1\}$
-\end_inset
-
- with 
-\begin_inset Formula $m$
-\end_inset
-
- subsets,
- and for the second one,
- we would need a partition of 
-\begin_inset Formula $\{1,\dots,n-1\}$
-\end_inset
-
- with 
-\begin_inset Formula $m-1$
-\end_inset
-
- subsets.
- Thus we would need
-\begin_inset Formula 
-\[
-\stirlb nm=m\stirlb{n-1}m+\stirlb{n-1}{m-1},
-\]
-
-\end_inset
-
-but this is precisely Equation 46.
-\end_layout
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-\begin_layout Standard
-\begin_inset ERT
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-\begin_layout Plain Layout
-
-
-\backslash
-rexerc67[M20]
-\end_layout
-
-\end_inset
-
-We often need to know that binomial coefficients aren't too large.
- Prove the easy-to-remember upper bound
-\begin_inset Formula 
-\begin{align*}
-\binom{n}{k} & \leq\left(\frac{n\text{e}}{k}\right)^{k}, & \text{when }n\geq k & \geq0.
-\end{align*}
-
-\end_inset
-
-
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-From Exercise 1.2.5–21,
- if 
-\begin_inset Formula $k\geq1$
-\end_inset
-
-,
- then 
-\begin_inset Formula $k!\geq\frac{k^{k}}{\text{e}^{k-1}}$
-\end_inset
-
-,
- so
-\begin_inset Formula 
-\[
-\binom{n}{k}=\frac{n(n-1)\cdots(n-k+1)}{k!}\leq\frac{n(n-1)\cdots(n-k+1)}{\frac{k^{k}}{\text{e}^{k-1}}}\leq\frac{n^{k}\text{e}^{k-1}}{k^{k}}\leq\left(\frac{n\text{e}}{k}\right)^{k}.
-\]
-
-\end_inset
-
-For 
-\begin_inset Formula $k=0$
-\end_inset
-
-,
-\begin_inset Formula 
-\[
-\lim_{k\to0}\left(\frac{n\text{e}}{k}\right)^{k}=\lim_{k\to\infty}\sqrt[k]{n\text{e}k}=1=\binom{n}{0}.
-\]
-
-\end_inset
-
-
-\end_layout
-
-\end_body
-\end_document