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\lhead{MATH 387-01}
\chead{Problem Set \#1\answer{ solutions}}
\rhead{\qonly{Name: \framebox{\phantom{Your name goes here}}}}
\lfoot{}
\cfoot{Page \thepage\ of \pageref{LastPage}}
\rfoot{{\bf due September 4, 2013}}
\begin{document}
\qonly{Show all your work, and explain why you use the arithmetic
operations you use in reaching an answer.}
\begin{enumerate}
\item {\bf (10 points)} \question{Let us call a number ``good'' if all
of its digits are the same parity (i.e. all odd or all even). So,
for instance, 62880 and 31713 are both good, but 61407 is not,
since 6, 4, and 0 have different parities from 1 and 7.}
\begin{enumerate}
\item {\bf (5 points)} \question{Determine the number of 3-digit
``good'' numbers; note that numbers, when conventionally
written, do not have a leading 0; e.g. ``024'' is not a
three-digit number.}
\answer{A ``good'' number might be either even or odd. If it is
even, there are four choices for the first digit (2, 4, 6, or 8)
and five for the second and third digits (0, 2, 4, 6, 8), so
there are $4\cdot 5\cdot 5=100$ even good three-digit numbers.
Likewise, an odd good number might have any of five different
possible choices for each of the three digits, so there are
$5\cdot 5\cdot 5=125$ odd good three-digit numbers.
Putting these two sets together, we see that there are
$100+125=225$ good three-digit numbers in all.}
\item {\bf (5 points)} \question{Find a formula in terms of $n$ for
the number of $n$-digit ``good'' numbers.}
\answer{The same logic can be applied as above, only instead of
having three digits, there are now $n$ digits; thus we can
compute that there are $4\cdot 5^{n-1}$ even good numbers and
$5^n$ odd good numbers for a total of $4\cdot 5^{n-1}+5^n=9\cdot
5^{n-1}$ good $n$-digit numbers.}
\end{enumerate}
\item {\bf (10 points)} \question{A modified deck of cards contains
five suits and ten cards (numbered 1--10) in each suit. A
``straight'' is defined as in poker as a hand in which the five
cards are numerically consecutive, and of any suit (ignore, for
the purposes of this problem, the poker convention that a hand of
this sort could also be a ``straight flush'' or ``royal flush''
instead). How many different five-card hands (which are not
ordered) are straights?}
\answer{The numbers on the cards must be one of six choices: 1--5,
2--6, 3--7, 4--8, 5--9, or 6--10. After we have chosen the
numbers, we choose a suit for each card; there are five choices
for each card, so in total we have $6\cdot 5^5=18750$ different
possible straights.}
\item {\bf (10 points)} \question{Let us consider license plates which
consist of two consecutive groups of either three letters (from a
pool of 26 possible) or three numbers (from a pool of 10
possible); e.g. 134-WRG, EEE-EEK, and 852-584 are all possible
license plates. However, to avoid confusion, you forbid either
group of 3 to consist entirely of the numbers ``0'', ``1'', and
``5'', or the letters ``O'', ``I'', and ``S'' (so we don't allow
115-RTG, because it might be misread as IIS-RTG). How many license
plates are possible according to this scheme (you may leave the
answer as an unsimplified arithmetic expression, if you like).}
\answer{Looking at a single group of three, there are $26^3$ ways to
build it with letters, and $10^3$ ways to build it with
numbers. Of those $(26^3+10^3)$ possibilities, we must reject
$3^3$ letter-sequences made with only O, I, and S, and $3^3$
number-sequences made with only 0, 1, and 5. Thus a single group
of 3 can be made in any of $(26^3+10^3-2\cdot 3^3)$ ways. Since we
build a license plate by selecting two such groups, there are
$(26^3+10^3-2\cdot 3^3)^2=345,997,201$ different license plates
possible.}
\item {\bf (10 points)} \question{Suppose we build a random
four-letter ``word'' by placing in order a random selection from
the 21 consonants, a random selection from the 5 vowels, another
random consonant, and then another random vowel (most words we
produce this way will be nonsense words like ``LEFE''). What is
the probability that we make a word in which all four letters are
different, like ``XOFA''?}
\answer{A probability when choosing randomly is simply the ratio
between the number of desirable results and the size of the entire
result pool. We can build an arbitrary ``word'' in any of $21\cdot
5\cdot 21\cdot 5=11025$ ways, and all of our letters can be
distinct in $21\cdot 5\cdot 20\cdot 4=8400$ ways, since for the
second consonant and vowel we must choose a different letter from
the one already chosen. Thus the probability is
$\frac{11025}{8400}=\frac{16}{21}\approx 76\%$.}
\end{enumerate}
\vfill
\begin{center}
\fbox{
\begin{minipage}{6in}
Musica est exercitium arithmeticae occultum nescientis se
numerare animi.\answer{ [Music is an arithmetical exercise of
the soul, which is unaware that it is counting.]}
\hfill ---Gottfried Leibniz
\end{minipage}
}
\end{center}
\end{document}