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Difficulty: HardCombinations and Group Selections

An executive board consisting of 8 distinct members needs to form a subcommittee of 4 members. However, board members Alex and Blair refuse to serve on the subcommittee together unless board member Morgan is also selected. How many different 4-member subcommittees can be formed under these conditions?

Answer: 60

Answer

60
The solution uses complementary counting. First, calculate the total possible 4-member subcommittees from 8 members without restrictions, which is (84)=70\binom{8}{4} = 70. Second, identify the restricted scenario that is not allowed: Alex and Blair are both selected, but Morgan is excluded. In this invalid scenario, 2 spots are taken by Alex and Blair, Morgan is excluded from consideration, leaving 2 spots to be filled from the remaining 5 board members, which equals (52)=10\binom{5}{2} = 10 invalid subcommittees. Subtracting the invalid subcommittees from the total gives 70−10=6070 - 10 = 60 valid subcommittees.

Step-by-Step Solution

1
Calculate the total number of ways to choose a 4-member subcommittee from 8 members without restrictions.
The total unrestricted combinations is (84)=8×7×6×54×3×2×1=70\binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70.
Since the order of selecting members into a subcommittee does not matter, use the combination formula (nk)\binom{n}{k}.
2
Determine the condition under which a subcommittee selection is invalid.
A subcommittee is invalid if and only if both Alex and Blair are selected AND Morgan is excluded.
Alex and Blair agree to serve together only if Morgan is also present. Thus, having Alex and Blair together without Morgan violates the condition.
3
Calculate the number of invalid subcommittees.
The number of invalid subcommittees is (52)=5×42×1=10\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10.
Alex and Blair take 2 of the 4 spots, and Morgan cannot take any spot. The remaining 2 spots must be filled from the remaining 8−3=58 - 3 = 5 members.
4
Subtract the invalid subcommittees from the total unrestricted subcommittees using complementary counting.
70−10=6070 - 10 = 60 valid subcommittees.
Complementary counting yields the total number of subcommittees that satisfy the restriction.

Key Concept

Combinations with Restrictions and Complementary Counting
Estimated Time:2m 0s
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Topics

Combinations and Group Selections

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