1.

Pizza Hut is offering customers opportunity to create for
themselves a unique music CD when they log on to the Pizza
Hut website and use an appropriate code number. The Pizza
Hut ad claims customers can make a 6song CD by selecting
from 200 different song titles.
(a) How many different sets of 6 songs could be selected
by a customer? (2
points)
We seek the number of 6song subsets from a set
of 200 songs, where order or arrangement is not
important. The solution is the number of
combinations of 6 items selected from 200, or
C(200,6)=(200!)/(194!*6!)=82,408,626,300.

(b) How many unique 6song arrangements could a customer
create? (2
points)
We seek the number of 6song arrangements from a
set of 200 songs, where order is important.
The solution is the number of permutations of 6
items selected from 200, or
P(200,6)=(200!)/(194!)=59,334,210,936,000.

Suppose that the 200 songs are listed in the following
categories: Pop (44), Rock & Roll (52), Rythem and Blues
(18), Jazz (21), Classical (36), and Show Tunes (29), with
each song listed in one and only one category. The numbers
in parentheses above indicate how many songs are listed in
each category.
(c) If a customer selects one song from each category,
how many sets of 6 songs could be selected by the customer?
(3 points)
We use the multiplication principle because we
want one from each category to create a 6song set.
The wording of the problem ("sets of 6 songs could
be selected") implies that order is not important.
The solution is 44*52*18*21*36*29=902,918,016.

(d) Pat refuses to consider any Rock & Roll titles for
her CD. How many 6song CD arrangements does Pat have to
choose from? (3
points)
We seek the number of 6song arrangements from a
set of only 148 songs (200 total songs less 52 R
& R songs). The wording of the problem ("6song
CD arrangements") implies that order is important.
The solution is the number of permutations of 6
items selected from 148, or
P(148,6)=(148!)/(142!)=9,484,150,515,840.


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2.

Three friends each have a red, a white, a yellow, a blue,
and a green Tshirt.
(a) If each chooses a shirt to wear, how many unique
3shirt sets could they be seen wearing?
(5 points)
We use the multiplication principle because we
want one shirt on each person in order to create a
3shirt set. There are 5 ways for each person to
choose a shirt, and the choice one person makes
does not alter the choice of anyone else. The
solution is 5*5*5=125 unique 3shirt sets.

(b) If each chooses a shirt to wear, how many ways are there
for each of them to all choose different colors?
(5 points)
We again use the multiplication principle
because we want one shirt on each person in order
to create a 3shirt set. This time, there are 5
ways for the first person to choose a shirt, 4 for
the next, and 3 for the third person. The number of
choice diminishes because the second and third
persons cannot choose a color that has already been
chosen. The solution is 5*4*3=60 unique 3shirt
sets with each shirt a different color.


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3.

Consider the letters in the word QUADRILLIONTHS.
(a) How many unique arrangements are there for the
letters in this word?
(2points)
There are 14! ways to arrange the 14 letters in
the word, but two of the letters appear twice each.
We account for the duplication of these letters by
dividing by 2!*2!. The solution is
(14!)/(2!2!1!1!1!1!1!1!1!1!1!1!)=(14!)/(2!2!)=21,794,572,800.

(b) How many arrangements exist if the fourletter sequence
QUAD must be kept together in the order it now appears?
(2 points)
Consider the chunk QUAD as one unit. We now have
11 units to permute.
There are 11! ways to arrange the 11 units in
the word, but two of the letters appear twice each.
We account for the duplication of these letters by
dividing by 2!*2!. The solution is
(11!)/(2!2!1!1!1!1!1!1!1!1!1!1!)=(11!)/(2!2!)=9,979,200.

(c) Considering only unique letters (no repetition), how
many 10letter subsets could be created from these letters?
(3 points)
There are 12 unique letters, so we seek the
number of 10letter subsets that can be created
from a set of 12 elements. Order is not important.
This is just a combination of 10 from 12, which is
C(12,10)=(12!)/(2!*10!)=66.

(d) How many arrangements exist if each must begin and end
with a vowel? (3
points)
Within the 14 places for letters, place the
first and last to assure they are vowels. There are
5 vowels to choose from for the first and 4 for the
last. This is P(5,2)=5*4.
Now place the remaining 12 letters. This can be
done in P(12,12)=12! ways.
Because each of the P(5,2) possible firstlast
vowel arrangements can be matched with the P(12,12)
ways to place the remaining letters, we multiple
the two results.
Finally, we account for the duplication of two
of the letters, requiring us to divide out those
equivalent groups. We divide by (2!*2!) as we did
in (a) and (b) above.
This results in
[P(5,2)*P(12,12)]/(2!*2!)=(5*4*12!)/(2!*2!)=2,395,008,000
ways to rearrange the letters with a vowel in the
first place and a vowel in the last place.


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4.

At the Westminister Kennel Club dog show, held this month
in Madison Square Garden in New York City and broadcast on
the USA channel, every entrant must be assigned a unique
registration number.
One suggested strategy for assigning registration numbers
requires a code that has three parts to it.
 The first digit ranges from 1 through 7,
corresponding to one of the seven dog groups the entrant
is classified in. These are the herding group, the
sporting group, the working group, the toy group, the
terrier group, the hound group, and the nonsporting
group.
 The next two digits correspond to the breed of the
dog within the particular group. An English Cocker
Spaniel, for instance, within the sporting group, is
breed 14. No group has more than 27 breeds.
 The final numbers in the registration code provide
more information about the particular dog. The first of
the final digits is a 0 or a 1, based on the sex of the
dog. The second and third digits in this last set
represent the age of the dog in years, starting with 00
if less than 9 months old and ranging up to 12, the
oldest dog in this year's show being 12 years old. The
last two digits in this final portion of the registration
code represent the state of residence of the dog. Values
01 through 50 represent the 50 United States. If a dog is
from outside the U.S., this number appears as 00.
(a) Without considering any other circumstances or
restrictions, how many unique registration codes are
possible under this scheme?
(8 points)
Consider the registration number and the number
of values that could occur in the various
positions. We then use the multiplication principle
to determine the total number of possibilities
because we want values in all the positions.
This results in 7*27*2*13*51=250,614 possible
registration codes.

(b) What problems in dog identification could occur under
this strategy? (2
points)
One problem is that two unique dogs in the show
may be assigned the same registration number. Two
or more dogs may have exactly the same
characteristics, as described by the different
criteria in the suggested registrationnumber
assignment scheme described above. There would be
no way, by this scheme, to distinguish between the
two dogs.


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5.

In a certain leap year, the 13th of the month was a
Friday three different times. What day of the week was 29
February that year? (10
points)
One way to approach this problem is to label the
days of the leap years using positive integers from
1 through 366.
The table below shows the labels for the 13th
day of each month for leap years. In parentheses
after a 13th's label is the remainder when the
label is divided by 7. Call this an index
number for the 13th of each month.
Month

LeapYear Labels and Index
Numbers for 13th of Each
Month

January

13 (6)

February

44 (2)

March

73 (3)

April

104 (6)

May

134 (1)

June

165 (4)

July

195 (6)

August

226 (2)

September

257 (5)

October

287 (0)

November

318 (3)

December

348 (5)

The only index number that occurs three times is
6. This means the index number 6 represents
Fridays.
 We determine which day of the week the 29th
of February lands in one of several ways. One
way is to move forward from the 13th of
February. This has index number 2. That means
the 20th of February has index number 2 as does
the 27th of February. That means the 29th of
February has index number 4. If index number 6
is Friday, index number 4 is two days earlier,
or Wednesday.
 Another way is to move back from the 13th of
March. The 13th of March has index number 3, so
the 6th of March and the 28th of February will
have index number 3. The 29th of February will
have index number 4, a Wednesday.
 A third way is to use the leapyear label
for the 29th of February. That label is 60 (31
days in January and 29 in February). We divide
60 by 7 and the remainder, 4, is the index
number for the 29th of February. As we have
already determined, index number 4 represents a
Wednesday.
When a leap year has three Friday the 13ths, the
29th of February will occur on a Wednesday.


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6.

Roberta claimed that the following equation was always
true for positive integers n > 1.
(a) Is Roberta correct?
(2 points)
(b) If Roberta is correct, justify her result for the
general case. If Roberta is not correct, provide an
example to show she is not correct (called a counter
example). (8
points)
To justify Roberta's claim in the general case,
we need to show that the left side and right side
of her equation are equivalent.
Here's one way to do that.
In the first line above, we have expanded n!
into its factors. In the second line, we rewrite
(n2)(n3)(n4)...(2)(1) in its equivalent form,
(n2)!. In the third line, we expand n(n1) to get
n^2n.
The third line is exactly the right side of
Roberta's equation. Therefore, we have shown how to
correctly and logically move from the left side of
Roberta's equation to the right side of Roberta's
equation. That shows us that the left side and
right side are equivalent, which in turn justifies
that the equation holds in general.


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