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\centerline{\bf Midterm Two, April 20 8:00PM-10:15PM on Zoom}
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\centerline{\bf WARNING: THIS MID IS THREE PAGES LONG!!!!!!!!!!!!!!!!!}


\begin{enumerate}

\item
(15 points)
\begin{enumerate}
\item
(5 points) 
What is the coefficient of $x^2y^3z^4$ in

$$(x+y+z)^9$$

\item
(10 points) 
What is the coefficient of $x^ay^bz^c$ in

$$(x+y+z)^{a+b+c}$$

\end{enumerate}


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\item
(20 points--5 points each)
The Narns play card games with a cards that have ranks in the set $\{1,2,\ldots,r\}$
and suites in the set $\{1,\ldots,s\}$. 
In Narn Poker, each player gets $h$ cards. 

We assume that both $r$ and $s$ are squares, so $\sqrt{r}$ and $\sqrt{s}$ are
natural numbers. 


\begin{enumerate}
\item
A {\it Square Hand} is a hand where all of the cards have square rank.

What is the probability of getting a Square Hand?

(Its okay if they are of the same suite, or not.) 

\item
A {\it Square Flush} is a hand where all of the cards have a square rank and all of the suites are the same. 

What is the probability of getting a Square Flush? 


\item
An {\it Apple} is when you get two of the same rank. There are no other restrictions,
so for example, if you had 3 of the same rank, that would still be an Apple. 

What is the probability of getting an Apple.


\end{enumerate}

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\item
(15 points) 
In this problem we guide you through the
birthday paradox with $m$ balls in $n$ boxes
where we want the probability that at least $k$ balls 
go in the same box is $\ge \frac{1}{2}$. 
(HINT: Follow the proof for THREE balls in a box
and feel free to use the approximations I use there.) 

Assume that $m$ is much less than $n$. 
Assume that $k$ is much less than both $n,m$.

We put $m$ balls into $n$ boxes at random. 

\begin{enumerate}
\item
Let $i_1,\ldots,i_k$ be $k$ balls. What is the probability they
are all in the same box?

\item
What is the (approx) probability that NO set of $k$ is in the same box?
(Use three approximations here: (a) that the events are independent, and
(b) use $(1-x)$ is approximately $e^{-x}$, and (c) $\binom{m}{k}\sim \frac{m^k}{k!}$. 

\item
Think of $n,k$ as being fixed but $m$ as being varrying.
Approximatly how large does $m$ have to be so that the prob
that $k$ are in the same box is $\ge \frac{1}{2}$? 

\end{enumerate}

\end{enumerate}
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