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\begin{document} 


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\begin{fminipage}
\centerline{\bf Counting}
\end{fminipage}




\begin{problem}
\begin{problemparts}
\problempart 
How many ways are there to color $n$ (distinguishable) balls 
using exactly $k$ colors? (Use no more and no fewer than $k$ colors.) 

\problempart 
If $x > k$ colors are available, how many ways are there
to color the $n$ balls using exactly $k$ of those colors?
\end{problemparts}
\end{problem}


\begin{problem}
How many permutations of ABCDEFG are there that don't contain
``$\cdots$BEG$\cdots$'' or ``$\cdots$AD$\cdots$''?
\end{problem}


\begin{problem}
Let $X$ be a set of size $n$, and let $S_1,S_2, \cdots, S_b$ be a collection
of (not necessarily distinct) subsets of $X$, all of size $l$. Suppose there
exist integers $t$ and $\lambda$ such that if $Y$ is any subset of $X$ of size
$t$, then $Y \subseteq S_i$ for exactly $\lambda$ of the sets $S_i$. (Of
course, $1 \leq t \leq l$ and $1 \leq \lambda \leq b$.) Suppose that $R$ is a
subset of $X$ of size $k$, for some $k$, where $1 \leq k \leq t$. Argue that
$R$ is a subset of exactly $\displaystyle \lambda { n-k \choose t-k }/
{ l-k \choose t-k }$ of the subsets $S_i$.
\end{problem}


\begin{problem}
I have $n$ one-dollar bills that I wish to donate  to $m$ charities.
\begin{problemparts}

\problempart 
In how many ways can I do this if I don't care how much each charity
gets? 

\solution{
Simply insert $m-1$ bars into a string of $n$ one-dollar bills. Then the
$i$-th charity gets the money between bars $i-1$ and $i$. This gives you 
${n+m-1 \choose m-1 } $ ways to distribute the money.
}

\problempart 
In how many ways can I do this so that each charity gets at
least~\$10?  

\solution{
First give 10 dollars to each charity, then distribute the remaining
 $n-10m$ dollars as in part a. This gives you ${n+m-1-10m \choose m-1}=
{n-9m-1 \choose m-1}$ ways as long as $n\geq 10m$.
}
\end{problemparts}
\end{problem}



\begin{problem}
In how many ways can $5$ indistinguishable covered wagons, $4$
indistinguishable uncovered wagons and $1$ horse be arranged in a
circle? 

\solution{
This problem is just a permutation of a multiset, except that we need
to account for rotations of a circle being the same. This can be
thought of either as fixing one object, say the horse, and arranging
the wagons around it (${9\choose 5}$ ways), or counting the permutations
and dividing by $10$ to avoid overcounting the rotations
($\frac{1}{10}\frac{10!}{5!4!1!}$). Either way the answer is 126. The
most common mistake was to forget about overcounting the rotations.
}
\end{problem}




\begin{problem}
How many $6$-digit decimal numbers are there that do not contain 
``$123$'' or ``$456$'' as a subsequence? (For purposes of this problem 
numbers may have leading $0$'s.)

\solution{
Let
\begin{itemize}
\item 
$A$ be the set of all six-digit strings, using the digits
$\{0, 1, \ldots, 9\}$.
\item 
$B$ be the set of six-digit strings that contain the subsequence
$123$.
\item 
$C$ be the set of all six-digit strings that contain the subsequence
$456$.
\item 
$D$ be the set of all six-digit strings that do not contain 
$123$ or $456$.
\end{itemize}
We're interested in figuring out $| D |$. Since 
$$
D = A - (B \cup C),
$$
and $B \cup C \subseteq A$,
we can use the inclusion-exclusion formula:
$$
| D | = | A | - |B \cup C| = | A | - | B | - | C | + | B \cap C |.
$$

First, $|A| = 10^6$, since we can put any of ten digits in each of
six places.  

Now, a string can contain $123$ in any of four positions; once this
position is chosen, we have $1000$ choices for the remaining positions.
However, this approach counts the string $123123$ twice, so 
the total size of $B$ is
$$
| B | = 4(1000) - 1 = 3999.
$$
By the same reasoning, we get
$$
| C | = 3999.
$$

Finally, there are only two strings in $B \cap C$, namely $123456$ 
and $456123$.  Thus,
$$
| B \cap C | = 2.
$$
Plugging this into our formula, we get
$$
| D | = 10^6  - 3999 - 3999 + 2 = 992004.
$$

{\em (Common Mistakes: A lot of people forgot about the double-counting
of $123123$ and $456456$, and so had $|B| = |C| = 4000$.  Some people also
argued that since $123$ can appear in one of only four places in the
string, $|B| = 4$.)}
}
\end{problem}





\begin{problem}
Find the number of three letter words over the alphabet a,b,\ldots,z 
that do not contain ``ab'' or ``ba''. 
\end{problem}




\begin{problem}
\begin{problemparts}
\problempart
How many ways can you arrange $39$ identical black balls and $13$
identical white balls in a row, so that no two white balls are
adjacent?
\problempart
A standard deck of $52$ cards is shuffled completely (i.e. any of the
permutations are equally likely.)  What is the {\bf probability} that
no two spades are adjacent in the shuffled deck?
\end{problemparts}
\end{problem}



\begin{problem}
A pitcher's record is $21$ wins and $4$ losses. {\bf Prove} that he
must have enjoyed a winning streak of $5$ games.
\end{problem}






\begin{problem}
Let $[n]$ denote the set $\{1,2,\cdots,n\}$.  Give formulas in $r$ and
$n$ for the number of elements in each of the sets defined below.
\begin{problemparts}

\problempart 
$
F(r,n) = |\{ f \mid f\mbox{ is a function from }[r] \mbox{ to }[n] \}|
$.

\solution{
\[F(r,n) = n^{r}.\]
}

\problempart 
$
J(r,n) = 
|\{ f \mid f\mbox{ is an injection from }[r] \mbox{ to }[n] \}|
$,
where $1 \leq r \leq n$.

\solution{
\[J(r,n) = P(n,r) = n!/(n-r)!.\]
To specify an injection we can choose among $n$ elements for $f(1)$,
$n-1$ elements for $f(2)$ (since we can't use the first), $n-2$ for
$f(3)$ and so on.  Thus the number of injections is
\[
n(n-1)(n-2) \cdots (n-r+1) = P(n,r)
\]
}

\problempart 
$S(r,n) = |\{ f \mid f\ \mbox{from}\ [r]\ \mbox{to}\ [n]\ \mbox{is
strictly increasing} \}|,\ \mbox{where}\ 1 \leq r \leq n$.

\solution{
\[
S(r,n) = C(n,r) = n!/r!(n-r)!.
\]
Since our function is strictly increasing, it can be completely specified by
the set of $r$ numbers that form the co-domain of $f$.  The smallest must be
$f(1)$, the second smallest $f(2)$ and so on.  Thus the number of strictly
increasing functions is $C(n,r)$.
}

\problempart
Order the following functions in terms of growth rate:
\[F(\sqrt{n}, n),\ F(n/2,n),\ F(n,n/2),\ J(n,n),\ S(n,n).\]

\solution{
Each function in the following list is $o(\cdot)$ of the next:
\[
S(n,n),\ F(\sqrt{n}, n),\ F(n/2,n),\ J(n,n),\ F(n,n/2).
\]
Recall that
\begin{eqnarray*}
S(n,n) & = & C(n,n) = 1  \\
F(\sqrt{n},n)& =& n^{\sqrt{n}}  \\
F(n/2,n) &=& n^{n/2} \\
J(n,n) &=& n! \\
F(n,n/2) &=& (n/2)^n
\end{eqnarray*}
To prove that the order above is indeed correct, we will use the fact
that if $f(n)$ and $g(n)$ are strictly positive functions then
\[
f(n) = o(g(n))
\quad\quad\text{iff}\quad\quad
\lim_{n \rightarrow \infty} \frac{f(n)}{g(n)} = 0
\]
$S(n,n)$ is clearly first on the list.  Now
\[
\frac{n^{\sqrt{n}}}{n^{n/2}} = n^{\sqrt{n} - n/2}
\]
and so as $n$ gets large, the ratio above approaches $0$.  
Note also that $n! \approx \sqrt{2\pi n}(n/e)^n$ and so the ratio
\[
\frac{n^{n/2}}{\sqrt{2\pi n}(n/e)^n} = \frac{\sqrt{n}^n}{\sqrt{2\pi
n}(n/e)^n} = \frac{1}{\sqrt{2\pi n}(\sqrt{n}/e)^n}
\]
approaches $0$ as $n$ increases.  Finally, 
\[
\frac{\sqrt{2\pi n}(n/e)^n}{(n/2)^n} = \frac{\sqrt{2\pi n}}{(e/2)^n}.
\]
Since $(e/2) > 1$, the denominator grows faster than any polynomial and so
that ratio goes to $0$.
}
\end{problemparts}
\end{problem}



\begin{problem}
A round pizza is divided into four equal slices (but still left on the
tray). Then nine pepperoni slices are placed on the pizza. 
Each pizza slice gets an integral number, possibly zero, of pepperoni
slices. 
\begin{problemparts}
\problempart
How many different ways are there to arrange the nine pepperoni slices
on the four (distinguishable) pizza slices?
% \problempart 
% Argue that there is no way to arrange the nine pepperoni slices on
% the pizza so that it has any sort of rotational symmetry. (That is,
% argue that rotation by any multiple of 90 degrees must give a
% different arrangement.)  What numbers, other than nine, have this
% property?

\solution{
We have to count the number of ways to arrange
9 indistinguishable objects (the pepperoni slices)
on four distinguishable pizza slices.
This is the same problem of distributing 
9 blank cards to 4 people,
and the answer is 
\[
BP(9,4) = { 9 + 4 - 1 \choose 4 - 1 } = {12\choose 3} = 220. 
\]
}


\problempart
How many different ways are there to arrange the nine pepperoni slices
on the four pizza slices, if two arrangements are considered ``the
same'' if one can be obtained from the other by a rotation by 
some multiple of 90 degrees?\dots

\solution{
We consider two arrangements to be the same if one 
can be obtained from the other by a rotation of the pizza.
Notice that if we take an arrangement 
and rotate 
the pizza by 0, 90, 180 or 270 degrees,
we obtain four distinct arrangements.
To see this, consider an arrangement,
that is a 4-tuple $(a,b,c,d)$ such that $a+b+c+d=9$,
and assume for contradiction that 
there are two rotations of $(a,b,c,d)$ that are the same.
This implies $a=c$ and $b=d$.
In particular, the sum 
$a+b+c+d=2(a+b)$ would be  even,
contradicting the hypothesis $a+b+c+d=9$.

So, the number of arrangements up to rotations of the pizza is
one fourth of the number of arrangements when 
considering the pizza slices to be indistinguishable,
and the answer is 
\[
{BP(9,4)\over 4} = {220\over 4} = 55. 
\]
}

\end{problemparts}
\end{problem}


\begin{problem}
A bridge hand consists of 13 cards.  The {\em suit-distribution}
of a hand is the number of cards in each suit.  For example, one
possible suit-distribution is 4 spades, 3 hearts, 3 diamonds, and 3
clubs.  
\begin{problemparts} 
\problempart 
How many different suit-distributions are there?

\problempart
How many different suit-distributions are there assuming that you get
at least one card in each suit?

\problempart 
Suppose the actual suits don't matter.  The {\em card-distribution} 
is specified by the {\em set} of numbers of each suit.  In the previous
example, the card-distribution would be $\{4, 3, 3, 3\}$.  How many
card-distributions are possible for a bridge hand?  (You need not give a
numerical answer, but may use notation from Handout 32 -- Notes on
Counting.)
\end{problemparts}
\end{problem}





\begin{problem}
Prove that among any set of 150 natural numbers, there must be three
numbers, $a$, $b$, and $c$, such that all the pairwise differences,
$a-b, a-c, b-c, b-a, c-a, c-b$, are multiples of 70.

\solution{
Use Pigeonhole. 
{\bf Pigeons} the 150 numbers.
{\bf Holes} the natural numbers mod $70$. 
There is some hole with $ \ceil{\frac{150}{70}} = 3$ pigeons, that is, 
three numbers $a,b,c$ that are equivalent modulo $70$. Then their
differences are equivalent to $0$ modulo $70$, therefore they are
divisible by $70$. 
}

\end{problem}





\begin{problem}
\begin{problemparts} 
\problempart 
Prove, by direct calculation, that the following
binomial identity is true for all $l,m,n$ such that $0 \leq l \leq m 
\leq n$ :
\begin{displaymath}
{n \choose m}{m  \choose l} = {n \choose l}{n-l \choose m-l}
\end{displaymath}
We just expand the binomial coefficients and cancel:
\begin{math}
{n \choose m}{n \choose l} = \frac{n!}{m!(n-m)!}\frac{m!}{l!(m-l)!} =
\frac{n!}{(n-m)!}\frac{1}{l!(m-l)!}=
\frac{n!}{l!(n-l)!}\frac{(n-l)!}{(m-l)!(n-m)!} = {n \choose l}{n-l
\choose m-l} 
\end{math}

\problempart 
Now prove the same identity combinatorially. Let $A$ be an
$n$-element set.  Explain how the left-hand and right-hand sides of the
identity in part a) may be interpreted as the number of ways of
choosing two subsets $B$ and $C$ of $A$, where $|B|=m$, $|C| =l$, and
$C \subseteq B$.

The left hand side corresponds to first choosing $B$, then $C$. The
number of ways of doing this is, using the product rule, ${n \choose
  m}{n \choose l}$. The right hand side corresponds to first choosing
$C$, and then choosing $B-C$. The number of ways of doing this is,
using the product rule, ${n \choose l}{n-l \choose m-l}$. Since both
sides are equivalent to the number of ways of choosing subsets $B$ and
$C$ such that $|B|=m$, $|C| =l$, and $C \subseteq B$, the identity
holds. 
\end{problemparts}
\end{problem}




\begin{problem}
Four distinct children approach you on halloween, requesting candy.
You have ten pieces of candy to distribute among them. In how many
different ways can you distribute the ten pieces if: (You must specify
your answer as an integer, or as a ratio of products of integers, or
as an integer raised to the power of an integer)\\ 
\begin{problemparts}

\problempart 
All the pieces are the same (Reese's Peanut-butter
Cups$^{(TM)}$). No other restrictions.
$$
{10+4-1 \choose 10} = {13 \choose 10} = \frac{13!}{3!10!}
$$

\problempart 
All the pieces are the same, and each child must receive at
least one piece. No other restrictions.
$$
{6+4-1 \choose 6} = {9 \choose 6} = \frac{9!}{3!6!}
$$

\problempart 
All the ten pieces are different. No other restrictions.
$$
4^{10}
$$

\problempart 
Five pieces are Reese's Peanut-Butter Cups $^{(TM)}$ and the other
five are Nestle's Crunch $^{(TM)}$. No other restrictions.
$$
{5+4-1 \choose 5}^2 = {8 \choose 5}^2 = (\frac{8!}{3!5!})^2
$$

\problempart 
All the pieces are the same. No child gets more than $4$
pieces. No other restrictions.
We use the formula for $10$-combinations of $4$ elements with
repetition with each element occuring fewer than $5$ times, and get:
$$
{13 \choose 3} - {4 \choose 1}{8 \choose 3} + {4 \choose 2}{3
  \choose 3} = 68
$$
\end{problemparts}
\end{problem}





\begin{problem}
\begin{problemparts}
\problempart
Six people communicate via radio frequency with one another. Each pair
of people choose one of {\bf two} available frequencies with which to
communicate.  Show that there is a subset of 3 people who are each
using the identical frequency with which to communicate with others in
the subset.

\problempart
Now suppose there are seventeen people communicating via radio 
frequency with one another. Each pair
of people choose one of {\bf three} available frequencies with which to
communicate.  Show that there is a subset of 3 people who are each
using the identical frequency with which to communicate with others in
the subset.

\end{problemparts}
\end{problem}


























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\newpage
\begin{fminipage}
\centerline{\bf Probability}
\end{fminipage}




\begin{problem}
In  $400$ rolls of a fair six-sided die, are you more likely to get
$220$ even numbers or to get $85$ sixes? Justify your answer.
\end{problem}



\begin{problem}
There is a bag with $999$ regular nickels and $1$ two-headed
nickel. Experiment X2$(k)$ consists of pulling a coin out of the bag at
random and then flipping the same coin $k$ times before putting it
back.
\begin{problemparts}
\problempart
You perform X2$(10)$. What is the probability you get 10 heads?

\solution{
Let $A$ be the event you draw the two-headed nickel and $B$
be the event you get 10 heads in a row. Then,
\begin{eqnarray*}
Pr(B) & = & Pr(B \wedge A) + Pr(B \wedge \overline{A}) \\
& = & Pr(B|A) Pr(A) + Pr(B|\overline{A})Pr(\overline{A}) 
\end{eqnarray*}
Clearly, $Pr(A) = 1/1000$ and $Pr(B) = 999/1000$.
If you get the two-headed nickel, you will get all
heads, so $Pr(B|A) = 1$.  The probability of getting ten heads with the
fair nickel is $2^{-10}$.  Thus, $Pr(B|\overline{A})=2^{-10}$.
Therefore, 
\[
Pr(B) = \frac{1}{1000} + \frac{999}{1000}2^{-10} =\frac{2023}{1024000}
\]
}

\problempart
You perform X2$(10)$. Given that you get $10$ heads what is the
probability that you pulled out the two-headed coin? 

\solution{
\[
Pr(A|B) = \frac{Pr(A \wedge B)}{Pr(B)}
\]
We have $Pr(B)$ from (a).  $Pr(A \wedge B) = 1/1000$, since
if we draw the two-headed nickel, we will get all heads. Thus,
\[
Pr(A|B) = \frac{1/1000}{2023/1024000} = \frac{1024}{2023}
\]
}

\problempart
You perform X2$(10)$. Given that you get $10$ heads what is the
probability that another $10$ flips with the coin you pulled out will
all come up heads?

\solution{
Let $C$ be the event that the next 10 flips are heads.
There are two ways to do this problem.

Method 1: 
Note that the event $C\wedge B$ is equivalent to getting 20 heads in a
row. 
\begin{eqnarray*}
Pr(C|B) &=& \frac{Pr(C\wedge B)}{Pr(B)} \\
&=& \frac{Pr(\mbox{20 heads} )}{Pr(\mbox{10 heads})} \\
&= & \frac{\frac{1}{1000} + \frac{999}{1000}2^{-20}}
	{\frac{1}{1000} + \frac{999}{1000}2^{-10}} \\
&=& \frac{1,049,575}{2,071,552}
\end{eqnarray*}

Method 2:
\begin{eqnarray*}
Pr(C|B) &=& Pr(C \wedge A|B) + Pr(C \wedge \overline{A}|B) \\
&=& Pr(C|A\wedge B)Pr(A|B) + Pr(C|\overline{A} \wedge B) Pr(\overline{A}|B) \\
\end{eqnarray*}

In (b), we determined $Pr(A|B)$.  $Pr(\overline{A}|B) = 1- Pr(A|B)$.

Now, $Pr(C|A\wedge B) = 1$, since the coin is two headed.
$Pr(C|\overline{A} \wedge B) = 2^{-10}$, since the flips
are independent.

Therefore,
\[
Pr(C|B) = \frac{1024}{2023} + \left(1-\frac{1024}{2023}\right) 2^{-10}
= \frac{1,049,575}{2,071,552}
\]
}
\end{problemparts}
\end{problem}




\begin{problem}
$10\%$ of the theorems in the lecture notes contain typos. For any
   theorem, it is the case that if you ask a TA whether it has a typo
   the probability that her answer is correct is $0.8$.
\begin{problemparts}
\problempart 
Given a theorem selected uniformly at random, if a TA tells you that
it has a typo, what is the probability that it does?	 

\problempart 
Suppose TAs answer independently from each other. If $2$ TAs tell you
a theorem has a typo, what is the probability that it does?
\end{problemparts}
\end{problem}


\begin{problem}
An integer $X$ is drawn from the set $1,2,\ldots,n$ uniformly at random. 
Prove that the probability that $X$ is divisible by $k$ is
$(1/n)(\lfloor n/k\rfloor)$.
\end{problem}







\begin{problem}
Mad Cow disease affects only 1 cow in a million.  A new test for the
disease always reports correctly when a cow is infected, but has a
probability $q$ of mistakenly reporting that an uninfected cow is
infected. 

Let $P(M|T)$ denote the probability that a randomly chosen cow has the
disease (event $M$) given that the test says the cow is infected
(event $T$).   
\begin{problemparts}
\problempart 
Write an expression for $P(M|T)$.  

\solution{
\begin{eqnarray*}
P(M|T)& =& \frac{P(M \wedge T)}{P(T)} \\
&=& \frac{P(T|M)P(M)}{P(T|M)P(M) + P(T|\overline{M})P(\overline{M})}\\
&=& \frac{(1)(10^{-6})}{(1)(10^{-6}) + (q)(1-10^{-6})}\\
&=& \frac{1}{1+999999q}
\end{eqnarray*}
}

\problempart 
How small does $q$ have to be for $P(M|T) \geq\frac{1}{2}$?

\solution{
For (b), $P(M|T)\geq \frac{1}{2}$ holds if 
\[
\frac{1}{1+999999q}\geq \frac{1}{2} \Longrightarrow q\leq \frac{1}{999999}.
\]
}

\end{problemparts}
\end{problem}





\begin{problem}
In a (five-card stud) poker game, two players $A$ and $B$ are dealt
$5$ cards each at random from the {\em same} standard deck of $52$.
%You may leave your answers in terms of the ratio of a product of
%integers and binomial coefficients.
\begin{problemparts}
\problempart
A {\em straight flush} consists of $5$ consecutive cards from the same
suit.  (Ace, 2, 3, 4, 5 is the lowest and 10, Jack, Queen, King, Ace
is the highest, so an Ace may not be in the middle of a straight
flush.)  What is the probability that player $A$ gets a straight
flush?


\problempart
A {\em four-of-a-kind} consists of $4$ cards of the same rank (for
example, four $7$s or four Kings).  The fifth card can be anything.
What is the probability that player $B$ gets a four-of-a-kind?


\problempart
What is the conditional probability that player $B$ gets a four-of-a-kind
{\em given} that player $A$ gets a straight flush?

\problempart 
What is the probability that player $A$ gets a straight flush {\em
and} player $B$ gets a four-of-a-kind?

\problempart
What is the probability that player $A$ gets a straight flush {\em
given} player $B$ gets a four-of-a-kind?

\end{problemparts}
\end{problem}



\begin{problem}
\begin{problemparts}
\problempart
At Marilyn's Casino, a new gambling game has been introduced.  
The game features 3 fair mutually independent 6-sided dice.  To play, the
contestant picks a number in the interval $[1,6]$.  The contestant
wins $\$1$ if his/her number comes up on one or more of the dice and loses
\$1 otherwise.  This game has proved to be very popular with the
gamblers such as Bobo who is a regular. Your job is to figure
out the probability of winning the game.
\vspace{2in}
\problempart
In an effort to attract even more gamblers to her casino, Marilyn is
thinking about a variation of the game in which the bettor also picks
a second chance number and the bettor wins if the first pick comes up
at least once or the second pick comes up at least twice. Compute the
probability of winning this game.
\end{problemparts}
\end{problem}




\begin{problem}
For a certain test for a newly discovered virus, called JABOLA, it is
known that $99\%$ of the people carrying the virus will test positive,
whereas only $1\%$ of the people not carrying the virus will test
positive. Suppose $0.5\%$ of the population carries the JABOLA virus.
Given that a person picked at random tests positive, what is the
conditional probability that this person carries the JABOLA virus?
\end{problem}


\end{document}                            
