FALL 2015
MATH 141 - ASSIGNMENT (6.1-6.4)
DUE: 10/19/15
Name
All steps must be written clearly and neatly to get full credit. If you use your calculator for
anything beyond an arithmetic calculation, please indicate how at the appropriate step.
1. (10pts) In a study of factors that influence brand switching by 18- to 24-year-old female
college students, the following factors were identified as being significant as judged by the
responses of the 439 participants in the survey:
1: Better price 2: Friend’s recommendation 3: Seeing others use it
4: Interesting packaging 5: Buzz - people talking about the brand
6: An advertisement 7: Press stories 8: Entertainer/sports celebrity endorsement
The results of the survey follow:
Factor
1
2
3
4
5
6
7
8
Respondents (%) 68.8 60.4 22.8 20.8 16.8 16.8 14.9 6.9
Let A denote the set of factors that had a response greater than 20%, let B denote the set
of factors that had a response between 15% and 25%, and let C denote the set of factors that
had a response less than 25%. Find
(i) n(A), n(B), n(C)
n(A) = 4, n(B) = 4, n(C) = 6. (Just count from the table.)
(ii) n(A ∩ B) by using the addition rule in the appropriate way
A ∪ B is the set of factors that had a response greater than 20%, or had a response
between 15% and 25%. From the table, there are 6 of them.
n(A∪B) = n(A)+n(B)−n(A∩B) so 6 = 4+4−n(A∩B), which means n(A∩B) = 2.
(iii) n(Ac ∩ C)
Ac ∩ C is the set of factors that had a response not greater than 20% and had a
response less than 25%. From the table, there are 4 of them. So n(Ac ∩ C) = 4.
(iv) n[(A ∪ B) ∩ C]
In (ii), we saw that there are 6 factors in A ∪ B, namely the first six factors. Out of
these six, the ones that are also in C are factors 3, 4, 5, and 6. So n[(A ∪ B) ∩ C] = 4.
You may also use a Venn diagram to answer various parts of this problem.
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2
ASSIGNMENT (6.1-6.4)
2. (20pts) Jack and Jill and 5 of their friends go to the movies. They all sit next to each
other in the same row. How many ways can this be done if
(i) There are no restrictions?
We are arranging 7 people on 7 seats. Either use the multiplication principle, which
gives us 7! = 5040, or use the calculator to compute P (7, 7) = 5040.
(ii) Jill must sit in the middle?
Use the multiplication principle. There is only 1 choice for the middle seat, then 6
choices for the first seat, 5 choices for the second, and so on.
6 · 5 · 4 · 1 · 3 · 2 · 1 = 720
(iii) Jill sits on one end of the row, and Jack sits on the other end of the row?
Jack and Jill occupy the two seats at the ends, and there are 2 ways for them to do
so - Jack left and Jill right, or the other way round.
Then the remaining 5 people can be arranged in the other 5 seats without restriction,
and there are 5! = 120 ways to do so.
The multiplication principle says that the total number of arrangements will be 2 ·
120 = 240.
(iv) Jack, Jill, or John sits in the middle seat?
There are 3 choices for the middle seat. Then the remaining 6 people can be arranged
without restriction, and there are 6! = 720 ways to do so.
The multiplication principle says that the total number of arrangements will be 3 ·
720 = 2160.
(v) Jack, Jill, and John sit together in the middle three seats?
Jack, Jill, and John will occupy the middle three seats, and there are 3! = 6 ways to
arrange them.
The remaining 4 people can be arranged in the remaining 4 seats, and there are
4! = 24 ways to do so.
The multiplication principle says that the total number of arrangements will be 6·24 =
144.
(vi) Jack and Jill must sit next to each other?
Since they must sit next to each other, we can first regard them as a single unit,
which means that we are effectively arranging 6 objects (the pair and the 5 friends),
and there are 6! = 720 ways to do so.
But Jack and Jill can swap seats, so there are 2 ways of arranging within the pair.
The multiplication principle says that the total number of arrangements will be 720 ·
2 = 1440.
(vii) Jill must not sit next to Jack?
We consider cases.
Case 1: Jill sits on either end
ASSIGNMENT (6.1-6.4)
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If Jill sits on the left end, then we get 1 · 5 · 5 · 4 · 3 · 2 · 1 = 600.
We get 5 on the second seat because Jack cannot sit there. From the third seat
onwards, the remaining 5 people can sit in any order.
Similarly, if Jill sits on the right end, then we get 1 · 2 · 3 · 4 · 5 · 5 · 1 = 600 if we go
from right to left.
Case 2: Jill does not sit on either end
Now Jill has 5 seats to choose from. Suppose she sits on the second seat from the
left. Then we get 5 · 1 · 4 · 4 · 3 · 2 · 1 = 480.
We have 1 on the second seat because Jill sits there. We have 5 on the first seat
because Jack cannot sit there, and for the same reason we have 4 on the third seat.
After those three seats have been occupied, the remaining 4 people can fill up the
remaining seats in any order.
By similar reasoning, we see that we get the same number, 480, if Jill sits on the
third seat, or the fourth, or the fifth, or the sixth.
Altogether, there will be 600 + 600 + 480 + 480 + 480 + 480 + 480 = 3600 ways.
3. (5pts) A student taking an examination is required to answer exactly 10 out of 15 questions.
(i) In how many ways can the 10 questions be selected?
(ii) In how many ways can the 10 questions be selected if exactly 2 of the first 3 questions
must be answered?
Solution:
(i) C(15, 10) = 3003
(ii) We choose 2 out of the first 3 questions. Then we choose 8 out of the remaining 12
questions.
C(3, 2) · C(12, 8) = 1485
4. (5pts) How many distinct ten-digit numbers can be made using the digits
1, 2, 2, 3, 3, 3, 4, 5, 6, 6?
Solution:
Some objects are identical. There are two 2s, three 3s, and two 6s.
10!
= 151, 200
2!3!2!