1.

Respond to each question below by placing your solution
in the blank. While you may show steps leading to your
solution, you do not need to generate written explanations
for questions (a) through (e) on this page.
a) How many distinct arrangements exist for the letters
in the word TRANSFERRERS? (2 points)
Solution: (12!)/(4!2!2!)
The numerator represents the number of ways to
arrange the letters in the given word. The
denominator represents the accounting for
duplication in arrangements due to duplicate
letters.

b) Consider the expansion of (k + j + m + n)^13.
(i) State the number of uncollected terms. (1
point)
(ii) State the value of the coefficient C in the
collected term Ck^2j^3m^5n^3. (1 point)
Solutions: (i) 4^13 (ii) (13!)/(2!3!5!3!)
(i) There are 13 opportunities to choose exactly
one letter from among the four k,j,m,n.
(ii) Imagine a 13letter word made up of 2
letters k, 3 letters j, 5 letters m, and 3 letters
n. Now follow the process used in Problem (1a)
above.

c) Use Pascal's Formula to express C(21,8) &endash;
C(20,7) as a single combination. (2 points)
Solution: C(20,8)
By Pascal's Formula, we know that
C(21,8)=C(20,8)+C(20,7). Now subtract C(20,7) from
each side to get the desired result.

d) Suppose that S(k) represents the sum of the elements
in the kth row of Pascal's Triangle. For instance, S(0) = 1
and S(1) = 2. Express S(9) + S(10) as a multiple of S(9). (2
points)
Solution: 3*S(9)
We know that the sum of the terms in row k is
2^k. Therefor, S(9)=2^9 and S(10)=2^10. The sum
S(9)+S(10)=2^9+2^10. We factor the right side to
get S(9)+S(10)=2^9(1+2)=3*2^9.

e) For the equation A+B+C+D+E+F=4, how many possible
solutions exist if the variables can take on values that are
nonnegative integers? (2 points)
Solution:C(9,5)
This problem is analogous to the problem of
determining the number of ways to distribute 4
objects among 6 categories, with the certainly the
one or more categories will have no
objects.



2.

Jack needed to arrange 12 hats on a display shelf. Seven
of the hats were red and the other five were blue. The hats
were distinguishable only by color.
a) If it was required that none of the blue hats could be
adjacent to one another, how many unique arrangements were
there of hats on the shelf? (5 points)
Solution: C(8,5) unique
arrangements
First, arrange the red hats. This can be done in
exactly one way, for they are indistinguishablefrom
one another. The seven red hats create 8 places
among them whereby blue hats can be places and have
blue hats remain nonadjacent. We therefore choose
5 of these 8 positions for the blue hats. Because
the blue hats are not distinguishable from one
another, we need account no further for their
arrangement.

b) Instead of (a), suppose it was required that all the
blue hats must be kept together. How many unique
arrangements of hats on the shelf are there under this
restriction? (5 points)
Solution: 8!/7! = 8 ways
Treat the blue hats as one unit. Together with
the 7 red hats, there are 8 objects to place.These
objects can be arranged in 8! ways, but we divide
by 7! to account for the 7 nondistinguishable red
hats.



3.

My inlaws live in a retirement community. Among all
individual residents within the community, we know
that::
 38 play golf
 21 play tennis
 56 play bridge
 8 play golf and tennis
 17 play golf and bridge
 13 play tennis and bridge
 5 play golf, tennis, and bridge
 72 do not play golf, tennis, nor bridge
a) How many individual residents are there in
this retirement community? (5 points)


Solution: 154 people
The Venn diagram here shows a picture of the
information provided. The eight numerical values in
the diagram represent all those in the community,
so the total population is just the sum of those
eight values.

b) How many of the individual residents participate in
only and exactly one of the three activities? (5 points)
Solution: 54 people
From the diagram, we see that 18 play only golf,
5 play only tennis, and 31 play only bridge. The
sum of these three values is the solution we
seek.



4.

Consider the letters in the word INSURRECTIONISTS.
a) How many unique arrangements are there for the letters
in this word? (3 points)
Solution: (16!)/(3!3!2!2!2!)
The numerator represents the number of ways to
arrange the letters in the given word. The
denominator represents the accounting for
duplication in arrangements due to duplicate
letters.

b) How many unique arrangements exist if each arrangement
must begin and end with the letter R? (3 points)
Solution: (14!)/(3!3!2!2!)
There are exactly 2 Rs, and because they are
identical, they can be placed in only one way.
There now remain 14 letters to place. The numerator
represents the number of ways to arrange those 14
letters. The denominator represents the accounting
for duplication in arrangements due to duplicate
letters.

c) How many unique arrangements exist if all the letters
I must be kept together and they may not appear at either
end of the word? (4 points)
Solution: (i) 14!/(3!2!2!2!)(2*13!)/(3!2!2!2!)
or (ii) (12*13!)/(3!2!2!2!)
The identical letters I can now be treated as
one unit. There remain 14 objects (13 letters and
the Igroup) to arrange.
(i) Arrange all 14 objects and subtract those
with the Igroup at either end.
(ii) Arrange the remaining 13 letters and place
the Igroup in one of the 12 slots among the 13
letters (not on the ends).
We can show that expressions (i) and (ii) are
identical.



5.

Iamso Piceune has a particular love for the
chocolatechip cookies served at Blaise's Bistro. Every
weekday, Monday through Friday, Iamso has coffee at the
Bistro and may or may not eat one or more chocolatechip
cookies.
a) Iamso decided he would consume exactly 15
chocolatechip cookies per week at the Bistro, and would
always eat at least one each day. Determine the maximum
number of weeks Iamso could do this without repeating the
same 5day cookieeating pattern. (5 points)
Solution: C(14,4) weeks
This is analogous to determining the number of
positive integer solutions to the equation
M+T+W+R+F=15. There are C(151,51) such
solutions.

b) For a long time, Iamso followed the cookieconsumption
pattern described in (a). Later, upon advice from his
doctor, he changed his cookieeating pattern. He decided to
eat only and exactly 5 cookies per week, with two
conditions:
 (i) It was okay, on one or more days of the week, to
eat no cookies, and
 (ii) He could never eat all 5 cookies in one
day.
How many different weekly cookieeating patterns could
Iamso follow under these conditions?
Solution: C(9,4)5 weekly patterns
By condition (i), this is analogous to solving
the equation M+T+W+R+F=5 over the nonnegative
integers. There are C(5+51,51)=C(9,4) such ways.
By (ii) we must eliminate from the C(i,4) ways
those with 5 as the value of one variable. There
are 5 such ways to eliminate.



6.

At a local opera, patrons can check their hats prior to
entering the auditorium.
a) Suppose that 6 people each check a hat prior to a
performance. Those same 6 people are now in line to retrieve
the hats. If the 6 hats are returned at random to the 6
people, with no attention paid to whether a hat is returned
to its rightful owner, what portion of all possible ways to
return the hats will result in no one receiving the correct
hat? Here's another way to pose this question: What is the
probability that no one receives a correct hat? (5
points)
Solution: D(6)/6!=265/720=53/144
(approximately 0.3681)
We need the ratio of the number of derangements
of 6 distinct objects to the total number of ways
to arrange 6 distinct objects.

b) Generalize the problem above to answer the question
for n people. What value does the probability approach as n
grows larger and larger? (5 points)
Solution:
D(n)/n! =
[n!(11+(1/2!)1/3!)+...+(1)^n*(1/n!)] /
n! = 11+(1/2!)1/3!)+...+(1)^n*(1/n!) which
approaches approximately 0.367879441171... or
exactly 1/e, where e is the natural
exponential.


