Math 15B
Midterm Exam Solutions
26 October 2001
1. Give an example of each of the following or explain why it cannot be done.
(a) A bijection from
{
1
,
2
,
3
,
4
}
to
{
a, b, c
}
.
A.
Impossible. A bijection has it’s range and domain the same size.
(b) A permutation of
{
1
,
2
,
3
,
4
,
5
,
6
,
7
}
that has a cycle of length 4 and also has a
cycle of length 5.
A.
Impossible. The sum of cycle lengths is the size of the set being permuted and
4 + 5
>
7.
2. A committee contains 7 women and 6 men. We want to form a subcommittee with 5
of these people.
(a) How many ways can this be done?
A.
(
7+6
5
)
(b) How many ways can this be done if the subcommittee must contain at least
2 women and at least 2 men?
A.
There are either (2 men AND 3 women) OR (3 men AND 2 women).
By the
Rules of Sum and Product, the answer is
(
6
2
)
×
(
7
3
)
+
(
6
3
)
×
(
7
2
)
.
Choosing two of each sex and then a fifth committee member via
(
6
2
)(
7
2
)(
9
1
)
over
counts. For example, if there are 3 women on the committee, the committee is
counted 3 times depending on which woman is selected as the fifth committee
member.
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 Fall '01
 Bender
 Math, Probability theory, Playing card, Cumulative distribution function, Suit, Card game

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