1.

Four boys and five girls, all of differing heights, stood
in line for a palm reading. If all the boys stand next to
each other in line, how many different linear arrangements
exist for the nine people?
Solution: 4!*6!
The four boys, standing next to each other, can
be arranged in P(4,4)=4! ways. Now treat them as
one unit, because they cannot be separated. This
gives us six units to permute for possible linear
arrangements, carried out in P(6,6)=6! ways.


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2.

A local fastfood outlet offered a variety of meal
combinations. Every meal combination included a sandwich, an
order of French fries, and a soft drink. Suppose there are 6
different sandwiches, 3 different sizes for French fries
orders, and 8 different soft drinks to choose from.
(a) How many meal combo orders must be placed to assure
that at least one meal combo is ordered twice? (4
points)
Solution: (6*3*8)+1=144+1=145
Use the multiplication principle to determine
there are 6*3*8=144 different combo meals. The
worst case scenario is that the first 144 orders
are each for different combo meals. By the
pigeonhole principle, the 145th order, therefore,
must assure us that at least one combo meal is
ordered twice.

The fastfood outlet also offers breakfast items. The
breakfast sandwiches include five different bagel
sandwiches, seven different biscuit sandwiches, and four
different English muffin sandwiches.
(b) Suzzie Softknuckle comes to the fastfood outlet for
a breakfast sandwich. How many different breakfast
sandwiches does she have to choose from? (3 points)
Solution: 16 choices
The sandwich types are disjoint from one
another, so use the addition principle to determine
there are 5+7+4 different sandwiches.

The Merchanteer County AllStars softball team stopped at
the fastfood outlet. Each of the 21 team members purchased
either a double cheeseburger or a hot hamandcheese
sandwich. No team member ordered more than one sandwich.
(c) Freddie the Fry Cook kept track of the sandwich
purchases in the exact order they were made. How many
different orderings were possible if k team members
purchased a hot hamandcheese sandwich? (3 points)
Solution: C(21,k)=C(21,21k)
In the list of 21 sandwich orders, select k of
them for hot hamandcheese. This can be done in
C(21,k) ways.


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3.

Five brothers each have the same set of seven hats,
distinguished only by color. Each has a white hat, a black
hat, an orange hat, a green hat, a yellow hat, a maroon hat,
and a red hat.
(a) If each brother chooses a hat to wear, how many 5hat
sets could they be seen wearing? (3 points)
Solution: 7^5
Each brother chooses one of seven hats. Each
brother's choice is independent of the others, so
by the multiplication principle, there are
7*7*7*7*7 different 5hat sets.

(b) How many ways are there for each of the brothers to
all choose hats of different colors? (3 points)
Solution: P(7,5)
The first brother has 7 colors to choose from,
the next has 6, and so on, down to the last brother
who has 3 colors to choose from.

(c) At a recent family reunion, four of the brothers were
seen wearing the same color hat while the fifth brother wore
a hat of a different color. In how many ways could this have
occurred? (4 points)
Solution: P(7,2)*C(5,1)=7*6*5
Choose one of the five brothers to wear the
differentcolored hat. This can be done in C(5,1)=5
ways. The set of four brothers choose one of seven
colors and the fifth chooses from the remaining six
colors.


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4.

4. Consider the letters in the word EXCESSIVENESS.
(a) How many unique arrangements are there for the
letters in this word? (2 points)
Solution: 13!/(4!1!1!4!1!1!1!)=13!/(4!4!)
We permute the 13 letters in the word, carried
out in 13! ways. We then divide by the denominator
shown here to account for the duplicate letters
that occur in the word.

(b) How many arrangements exist if the threeletter
sequence EXC must be kept together in the order shown? (2
points)
Solution: 11!/(1!4!4!1!1!1!)
Treat EXC as a unit, resulting in 11 objects to
permute. We now carry out the same process as in
(a) above.

(c) How many arrangements exist if each must begin and
end with a consonant? (3 points)
Solution:
[P(8,2)*P(11,11)]/(4!*4!)=(8*7*11!)/(4!*4!)
Within the 13 places for letters, place the
first and last to assure they are consonants. There
are 8 consonants to choose from for the first and 7
for the last. This is P(8,2)=8*7. Now place the
remaining 11 letters. This can be done in
P(11,11)=11! ways. Because each of the P(8,2)
possible firstlast consonant arrangements can be
matched with the P(11,11) ways to place the
remaining letters, we multiple the two results.
Finally, we account for the duplication of letters,
requiring us to divide out those equivalent groups.
We divide by (4!*4!) as we did in (a) above. This
results in
[P(8,2)*P(11,11)]/(4!*4!)=(8*7*11!)/(4!*4!)
ways to rearrange the letters with a consonant in
the first place and a consonant in the last
place.

(d) How many 3letter sets can be created using only the
unique letters in the word? (3 points)
Solution: C(7,3)
There are 7 unique letters in the word and we
select 3 of them.


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5.

At Bunion College, a small undergraduate institution in
the midwest, every student must select a password to use to
enter the college computer network. In creating a password,
the following restrictions must all be met.
 The password must contain only digits or letters of
the alphabet.
 The password must be four or five characters in
length.
 The password must begin with either a B or a C.
How many different passwords are possible under these
restrictions?
Solution: 2*36^3 + 2*36^4
Consider two cases, one for the 4character
password and one for the 5character password.
Case I: For a 4character password, there are 2
choices for the first position (B or C) and 36
choices for each of the next three positions,
resulting in 2*36^3 different 4character
passwords.
Case I: For a 5character password, there are
again 2 choices for the first position (B or C) and
36 choices for each of the next four positions,
resulting in 2*36^4 different 5character
passwords.
Because the cases are disjoint&emdash;we cannot
have a password that is both 4 and 5 characters
long&emdash;we can add the results of the two
cases.


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6.

Francisco approached his combinatorics instructor and
showed her the following claims:
(i) 5! + 5! = 10! (ii) 2! &endash; 1! = 1! (iii) 5!
= 5*4!
(a) State whether each claim is true or false. Show
arithmetic work to support your response. (3 points)
Solution
(i) 5! + 5! = 10! False: 5!=120, so 5!+5!=240
but 10!=3,628,800.
(ii) 2!  1! = 1! True: 2!=2 and 1!=1, so 21=1
is true.
(iii) 5! = 5*4! True: 5!=5*4*3*2*1, but
4*3*2*1=4!, so 5!=5*4!.

(b) For any of the three claims that are true, write a
generalization of Francisco's claim. (3 points)
(ii) 2!  1! = 1!

Generalization: n!(n1)!=(n1)! Or,
alternatively, n!1!=1!

(iii) 5! = 5*4!

Generalization: n!=n*(n1)!


(c) For each generalization you wrote for (b), determine
whether it is true or false. If it is false, provide a
counterexample to show that; if it is true, justify that the
result holds in general. (4 points)
(iia) n!(n1)!=(n1)!

False: Counterexample: 6!5! is not
equal to 5!

(iib) n!1!=1!

False: Counterexample: 10!1! is not
equal to 1!

(iii) n!=n*(n1)!

True: By the definition of factorials
this must always hold.



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