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\centerline{\bf Math 365 -- Wednesday 2/27/18 - 6.4: Binomial coefficients and identities}
\noindent Recall
$$\binom{n}{k} = \frac{n!}{(n-k)! k!} .$$
Tips for being slick:
\begin{enumerate}
\item The values like
$\binom{n}{0} = 1, \quad \binom{n}{1} = n, \quad \binom{n}{n} = 1$
are totally straightforward, and should not involve writing down formulas with factorials. In fact, you're more likely to make mistakes with the algebra than if you step back and ask yourself ``How many ways can I pick one thing from 5?'', ``\dots nothing from 5?'', ``\dots everything from 5?''
\item When doing computations, it's often easier to pre-cancel $(n-k)!$ or $k!$ from $n!$.
For example,
$$\binom{11}{3} = \frac{(11!/(11-3)!)}{3!} = \frac{11*10*9}{3*2*1} = 11*5*3= 165.$$
However, if $k$ is bigger than half of $n$, then use $k!$ to cancel terms from $n!$ instead. For example,
$$\binom{11}{8} = \frac{(11!/8!)}{(11-8)!}= \frac{11*10*9}{3*2*1} = 165,$$
rather than writing out, say,
$$\binom{11}{8} = \frac{(11!/(11-8)!)}{8!} = \frac{11*10*9*8*7*6*5*4}{8*7*6*5*4*3*2*1}.$$
\end{enumerate}
\begin{warmup}
\underline{Without a calculator}, compute the following values:
{
$$\begin{array}{l@{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}l@{\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}
\displaystyle\binom{4}{2}=&
\displaystyle\binom{5}{3}= \\~\\~\\~\\~\\
\displaystyle \binom{8}{6} = &
\displaystyle\binom{10}{3} = \\~\\~\\~\\~\\
\displaystyle\binom{10}{9} = &
\displaystyle\binom{15}{0} = \\~\\~\\~\\~\\
\displaystyle\binom{200}{1} = &
\displaystyle\binom {200}{200} =
\end{array}$$
}
\pagebreak
\end{warmup}
\begin{try}~
\begin{enumerate}[(a)]
\item Expand $(x+y)^5$ and $(x+y)^8$ using the binomial theorem.
\item Substitute $x = 2z$ and $y = 3$ to calculate $(2z + 3)^4$.
\item What identity do you get if you substitute $x = -1$ and $y = 1$ in the binomial theorem?
\\
Check your identity (like in the previous problem) for $n = 4$.
\item Using the binomial theorem to prove combinatorial identities.
\begin{enumerate}[(i)]
\item Use the binomial theorem to explain why
$$2^n = \sum_{k=0}^n \binom{n}{k}$$
Then check and examples of this identity by calculating both sides for $n=4$. \\
(Hint: substitute $x = y = 1$).
\item Use the binomial theorem to explain why
$$2^n = (-1)^n\sum_{k=0}^n \binom{n}{k}(-3)^k.$$
Then check and examples of this identity by calculating both sides for $n=4$. \\
(Hint: what other examples can you think of of integers that sum to 2?).
\end{enumerate}
\item Give a formula for the coefficient of $x^k$ in the expansion
of $(x + 1/x )^{100}$ , where $k$ is an integer.
\item As a useful counting tool, we have (so far) only defined $\binom{n}{k}$ for non-negative integers $n$ and $k$. But in \S8.4 (p.\ 539, Def.\ 2), the book defines ``extended binomial coefficients'' as follows: \\
Let $u \in \RR$ and $k \in \ZZ_{\ge 0}$. The \emph{extended binomial coefficient} $\binom{u}{k}$ is defined by
$$
\binom{u}{k} = \begin{cases} u(u-1)(u-2) \cdots (u-k+1)/k! & \text{if } k > 0\\ 1 & \text{if } k=0.\end{cases}
$$
For example,
$$\binom{0.2}{3} = \frac{0.2(-0.8)(-1.8)}{3*2*1}.$$
\begin{enumerate}[(i)]
\item Compute $\binom{\pi}{4}$, $\binom{1/2}{2}$, and $\binom{7/3}{0}$.
\item Verify algebraically that if $n$ is a positive integer and $0 \le k \le n$, then
$$\binom{-n}{k} = (-1)^k \binom{n+k-1}{k}.$$
Check this identity for $n=5$ and $k=3$ (you should probably do this first).
\item BONUS: The \emph{extended binomial theorem} states that for any real number $u$, we have
$$(1+x)^{u} = \sum_{i=0}^\infty \binom{u}{k} x^k.$$
Now recall from calculus the Taylor series expansion
$$(1+x)^{-1} = \sum_{i=0}^\infty x^i(-1)^i.$$
Check that the first 3 terms ($i=0,1,2$) of our known Taylor series expansion match the first 3 terms of the extended binomial theorem expansion (when $u = -1$). Finally, verify that this example matches correctly for all terms by showing that $\binom{-1}{k} = (-1)^k$ for any $k \in \ZZ_{\ge0}$.
\item BONUS: Show in that the Taylor series expansion for $(1+x)^{-n}$ matches the extended binomial theorem for $n=3$.
\end{enumerate}
\end{enumerate}
\end{try}
\pagebreak
\begin{try}~
\begin{enumerate}[(a)]
\item Explain the example provided for the proof of Vandermonde's in the notes using words.
\item Substitute $m = r = n$ into Vandermonde's identity to show that
$$\binom{2n}{n} = \sum_{k=0}^n \binom{n}{k}^2,$$
and check this identity for $n = 2$.
\item Consider the identity
$$\binom{n}{k} k = \binom{n-1}{k-1} n$$
for integers $1 \leq k \leq n$.
\begin{enumerate}[(i)]
\item Verify this identity for $n = 5$ and $k = 3$.
\item Explain why this identity is true using a combinatorial argument.\\
{[}\emph{Hint: Count, in two different ways, the number of ways to pick a subset with $k$ elements from a set with $n$ elements, along with a distinguished element of that $k$-element subset. For example, out of $n$ people, pick a committee of $k$ people and choose someone on that committee to organize their meetings.
}{]}
\item Illustrate your combinatorial proof using the set $A = \{a,b,c\}$ (so that $n=3$) and $k=2$.
\item Verify the identity algebraically using the formula $\binom{n}{k} = n!/((n-k)!k!)$.
\end{enumerate}
\end{enumerate}
\end{try}
\vfill
\begin{try}~
\begin{enumerate}[(a)]
\item Consider strings of length 10 consisting of 1's, 2's, and/or 3's.
\begin{enumerate}[(i)]
\item How many of these are there?
\item How many of these are there that contain exactly three 1's, two 2's, and five 3's?
\item How many of these are there that contain at least three 1's, and exacly four 2's?
\end{enumerate}
\item How many anagrams are there of MISSISSIPPI?
\end{enumerate}
\end{try}
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