1.

A bus filled with a high school music group stopped at
Blaise's Bistro. The director, a budding mathematician,
noticed the menu board at the Bistro and quickly assured her
assistant that at least one of the 8 sodas listed on the
board would be ordered at least 4 times by the student
musicians. Naturally, the director assumed that each student
would order exactly one soda from the list.
What is the minimum number of students that must be in
this music group?
Suppose that none of the 8 sodas is ordered four
times or more. It is possible that all 8 varieties
were ordered three times each, requiring 24
students. A 25th student, however, would force the
fourth order of one of the sodas, by the Pigeonhole
Principle. Therefore, the minimum number of
students in the group is 25.


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2.

North American radio stations must adhere to specific
guidelines when selecting the call letters for the station
name.
 The name must contain either three or four letters of
the alphabet.
 The name must begin with a W or a K.
How many different radio station names are possible under
these restrictions?
Consider two disjoint cases.
Case I: 3 letters in the name
In this case, there are 2 choices for the first
letter and 26 choices for each of the second and
third letters. Each choice is independent so by the
multiplication principle, there are 2*26*26
possible 3letter names.
Case II: 4 letters in the name
In this case, there are 2 choices for the first
letter and 26 choices for each of the second,
third, and fourth letters. Each choice is
independent so by the multiplication principle,
there are 2*26*26*26 possible 4letter names.
Because the cases are disjoint, we add the
results of the two cases to determine the total
number of names that are possible.
2*26*26 +
2*26*26*26 = 1352 + 35,152 = 36,504


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3.

Consider the letters in the word SIMULATE.
 (a) How many rearrangements are there of these
letters?


 (b) How many rearrangements exist if the threeletter
sequence SIM must be kept together as it appears?

Consider the chunk SIM as one unit. We now
have 6 units to permute. This is just
P(6,6)=6!

 (c) How many rearrangements exist if each must begin
and end with a vowel?

Within the 8 places for letters, place the
first and last to assure they are vowels. There
are 4 vowels to choose from for the first and 3
for the last. This is P(4,2).
Now place the remaining 6 letters. This can
be done in P(6,6) ways.
Because each of the possible firstlast vowel
arrangements can be matched with the P(6,6) ways
to place the remaining letters, we multiple the
two results.
This results in P(4,2)*P(6,6) ways to
rearrange the letters with a vowel in the first
place and a vowel in the last place.

 (d) From the 8letter set, how many 5letter subsets
exist?

This is just a combination of 5 from 8:
C(8,5).


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4.

Using the set {A,B,C}, what fraction of all 5letter
words that can be created contain exactly one A?
There are a total of 3*3*3*3*3 = 243 5letter
words that can be made.
To determine the number of words with exactly
one A, we note that there are 5 places for the one
A. That is, an A can and must be placed in one of
the five positions in the word. There remain 4
places and for each there are two choices. This
results in 5*2*2*2*2 or 80 possible 5letter words
that contain exactly one A.
This results in 80/243 as the fraction of all
5letter words, from the set {A,B,C}, that contain
exactly one A.


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5.

Al and Bobbie are in a group of 12 students. Three teams
of 4 students are to be created from the group of 12. Among
all possible threeteam sets, how many ways exist for Al and
Bobbie to be on different teams?
There are 3*2 = 6 ways to place Al (A)
and Bobbie (B) onto two different teams from the
three teams being created. Although not necessary,
you can think of the teams with names RED, BLUE,
and GREEN, and count the possible ways A and B
could be placed onto two different teams.
Now that each of A and B are in a designated
team, choose from the remaining students to fill
out the teams. From the 10 remaining students,
choose 3 to fill out the team A is on. This can be
done in C(10,3) ways. Now from the 7 remaining
students, choose 3 to fill out the team B is on.
This can be done in C(7,3) ways. Finally, place the
remaining 4 students on the third team, the one
that neither A nor B is on.
This results in 6*C(10,3)*C(7,3) ways to meet
the conditions of the problem.


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6.

Complete ONE of the two problems listed below.
I. Show an algebraic proof that
C(a,c)*C(ac,bc) = C(a,b)*C(b,c) for
a>=b>=c.
The strategy to use here is to represent each
combination expression using factorial notation and
through simplification, show that the leftside
product is equivalent to the rightside
product.
The final expressions in [1] and
[2] above are equivalent, thereby showing
that C(a,c)*C(ac,bc) = C(a,b)*C(b,c).
This completes the required algebraic proof.

II. Determine the smallest natural number n that
assures
We have shown in class that ,
so our goal is to determine the smallest natural
number n so that . We first manipulate the
inequality:
Using conventional methods for solving equations
(complete the square or use the quadratic formula
on [4], carry out a guessandtest on
[3], graph the expression on the LS of
[4] and look for xaxis intercepts, use a
CAS SOLVE routine,
and so on), we determine that the required natural
number n is 1414.


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