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Difficulty: Very hardPermutations, Combinations, and Fundamental Counting Principle

A committee of 88 people consists of 44 men and 44 women. A subcommittee of 44 people is to be selected from this group such that the subcommittee contains at least one man and at least one woman. If two specific members, one man and one woman, refuse to serve together on the same subcommittee, how many different valid subcommittees of 44 people can be formed?

Answer: 53

Answer

53
The total number of ways to choose 4 people out of 8 is (84)=70\binom{8}{4} = 70. Removing the 2 single-gender subcommittees (4 men or 4 women) leaves 68 gender-valid subcommittees. Among these 68 subcommittees, exactly (62)=15\binom{6}{2} = 15 contain both of the two conflicting individuals. Subtracting these 15 forbidden subcommittees gives 68−15=5368 - 15 = 53 valid subcommittees.

Step-by-Step Solution

1
Calculate total ways to pick 4 people out of 8 without restrictions
\binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70
This establishes the total baseline sample space of possible 4-person groups.
2
Exclude single-gender groups to satisfy the gender balance constraint
70 - \binom{4}{4} - \binom{4}{4} = 70 - 1 - 1 = 68
Groups with 0 men or 0 women are invalid.
3
Count the forbidden groups that contain both of the conflicting individuals
\binom{6}{2} = 15
Fixing the 2 specific individuals in the subcommittee requires selecting 2 additional members from the remaining 6 people.
4
Subtract forbidden groups from gender-valid groups
68 - 15 = 53
Every group containing both conflicting individuals already satisfies the gender constraint, so exactly 15 invalid groups must be removed from the 68 gender-valid groups.

Key Concept

Combinations with multiple overlapping constraints (complementary counting)
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Topics

Permutations, Combinations, and Fundamental Counting Principle

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