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Q17
Q919
Q21
CHAPTER 5
Section 5.3, pp. 286–288
Solutions to Odd Number Problems
1.
For each coin, you can choose to include it or choose not to include it. Assuming
that you will include at least one coin,
number of sums = 2
4
– 1
= 15
3.
a)
n
= 2
2
= 4
b)
n
= 2
4
= 16
c)
n
= 2
7
= 128
5.
a)
Combinations; order is unimportant.
12
C
3
= 220
b)
Permutations; order is important.
12
P
3
= 1320
c)
Combinations; order is unimportant.
(
3
C
1
) (
4
C
1
) (
2
C
1
) = 24
d)
Permutations; order is important.
4
P
2
= 12
7.
There are 3 chocolatechip cookies, 2 peanutbutter cookies, 1 lemon cookie, 1
almond cookie, and 5 raisin cookies in the jar. You may choose all, some, or none
of each type of cookie. Assume that you will take at least one cookie.
a)
n
= (4)(3)(2)(2)(6) – 1
= 287
b)
(3)(3)(2)(2)(6) = 216
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View Full Document9.
There are 24 cards to choose from.
a)
There are 12 red cards.
n
(5 red cards) =
12
C
5
= 792
b)
n
(at least two red cards) =
n
(no restrictions) –
n
(no red cards) –
n
(1 red card)
=
24
C
5
– (
12
C
0
)(
12
C
5
) – (
12
C
1
)(
12
C
4
)
= 35 772
c)
n
(at most 2 red cards)
= (
12
C
0
)(
12
C
5
) + (
12
C
1
)(
12
C
4
) + (
12
C
2
)(
12
C
3
)
= 21 252
11.
5880 = (2
3
)(7
2
)(3)(5)
a)
Total number of divisors including 1 is (4)(3)(2)(2) = 48.
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 Spring '10
 Mr.M

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