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\noindent
{\bf Solutions to HW 10.}

\bigskip
  
\begin{enumerate}
\item[1.] This problem involves a deck of $52$ distinct playing cards.
\begin{enumerate}
\item[(a)]
  In how many ways can a
  $13$-card bridge hand be dealt from the deck?
  \bigskip
  
\color{blue}
We want an ordered sequence of $13$ cards chosen from a $52$-element set.
There are $13!\cdot \binom{52}{13}=(52)_{13}=\frac{52!}{39!}$ of these.
 \color{black}
\bigskip

\item[(b)]
  How many different $13$-card bridge hands are there?
\end{enumerate}  
\bigskip

\color{blue}
We want a $13$-card subset of a $52$-element set.
There are $\binom{52}{13}=\frac{52!}{13!39!}$ of these.
\color{black}
\end{enumerate}
\bigskip

\begin{enumerate}
\item[2.]
\begin{enumerate}
\item[(a)] How many paths are there from the point $(0,0)$
of $\mathbb R^2$
to the point $(10,15)$ of $\mathbb R^2$ if each path 
consists of a sequence of steps of length 1 moving 
in the direction of the positive $x$-axis or the positive $y$-axis?
\bigskip

\color{blue}
We can describe a path by a list of instructions of the form
``$(x,x,y,x,y,y,\ldots, x,y,y)$'', which is a string of $10$ $x$'s and
$15$ $y$'s in some order. If the first two instructions are $x$,
then we take our first two steps in the $x$ direction;
if our next instruction
is $y$, we take our next step in the $y$ direction. ETC.

The number of paths is equal to the number
of lists of instructions, which is equal to the number
of strings of length $10+15=25$ consisting of $x$'s and $y$'s,
which contain $10$ $x$'s and
$15$ $y$'s.
This number is $\binom{25}{10} = \frac{25}{10!15!}$.
(You have 25 instructions, and you ``choose''
10 instructions to be $x$'s and leave the remaining
15 to be $y$'s.)
\color{black}

\bigskip

\item[(b)] How many paths are there from the point $(0,0,0)$ 
of $\mathbb R^3$
to the point $(10,15,20)$ of $\mathbb R^3$ if each path 
consists of a sequence of steps of length 1 moving 
in the direction of the positive $x$-axis, the positive $y$-axis
or the positive $z$-axis?
\bigskip

\color{blue}
Using the same reasoning, we want to count strings of length
$10+15+20=45$ which have $10$ $x$'s, $15$ $y$'s, and $20$ $z$'s.
The number is $\binom{45}{10,15,20} = \frac{45!}{10!15!20!}$.
\color{black}
\end{enumerate}
\end{enumerate}
\bigskip
  
\begin{enumerate}
\item[3.]  
Let $MC(n,k)$ be the number ``$n$-multichoose-$k$''.
  Use a combinatorial
argument to show that
$MC(n,0)+MC(n,1)+\cdots+MC(n,k)=MC(n+1,k)$.
\end{enumerate}

\bigskip

\color{blue}
      {\bf Solution 1.} (A combinatorial argument.)
      Let $D$ be the set of all distributions of
      $k$ identical balls to $n+1$ distinct boxes
      with repetition allowed.
      $|D| = MC(n+1,k) = \left(\binom{n+1}{k}\right)=\binom{n+k}{k}$.

      Partition $D$ into sets $D_0, D_1, \ldots, D_k$ where
      $D_i\subseteq D$ is the number of distributions
      where $i$ balls are distributed to the first $n$ boxes,
      while the remaining $k-i$ balls are distributed to the last box.
      Since we can distribute $i$ identical balls to
      the first $n$ boxes in 
      $MC(n,i)$ ways, and we have no choice but to put the
      remaining $k-i$ balls into the last box,
      we have $|D_i| = MC(n,i)$. Since we have a partition,
      \[MC(n+1,k)=|D|=|D_0|+|D_1|+\cdots+|D_k|=MC(n,0)+MC(n,1)+\cdots+MC(n,k).\]
      
\bigskip      
    {\bf Solution 2.} (Not a combinatorial argument, but OK.)
    \[
    \begin{array}{rl}
MC(n,0)+MC(n,1)+\cdots+MC(n,k)&=      
\binom{n-1}{0}+\binom{n}{1}+\binom{n+1}{2}+\cdots+\binom{n+k-1}{k}\\
&\vphantom{A}\\
&=[\binom{\color{red}n\color{blue}}{0}+\binom{n}{1}]
+\binom{n+1}{2}+\cdots+\binom{n+k-1}{k}\\
&\vphantom{A}\\
&=[\color{red}\binom{n+1}{1}\color{blue}]+\binom{n+1}{2}+
\cdots+\binom{n+k-1}{k}\\
&\vphantom{A}\\
&=[\binom{n+1}{1}+\binom{n+1}{2}]+\cdots+\binom{n+k-1}{k}\\
&\vphantom{A}\\
&\mbox{}\hphantom{AAAAA}\vdots\\
&\vphantom{A}\\
&=\binom{n+k-1}{k-1}+\binom{n+k-1}{k}\\
&\vphantom{A}\\
&=\binom{n+k}{k} = MC(n+1,k).
    \end{array}
    \]
\color{black}

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