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Difficulty: Very hardPermutations and Combinations

A financial advisory board of 66 members is to be selected from a pool of 66 senior analysts and 55 junior analysts. The board must include at least 22 senior analysts and at least 22 junior analysts. However, two specific senior analysts, AA and BB, cannot both serve on the board together. How many different 66-member boards can be formed under these conditions?

  1. A
    300
  2. 320Answer
  3. C
    325
  4. D
    336
  5. E
    425

Answer

The total number of valid 6-member boards that can be formed is 320.
To find the number of valid boards, we first calculate all possible 6-member boards satisfying the minimum criteria of having at least 2 senior analysts and at least 2 junior analysts. The possible (senior, junior) distributions are (4,2), (3,3), and (2,4). Calculating each case gives (64)(52)=150\binom{6}{4}\binom{5}{2} = 150, (63)(53)=200\binom{6}{3}\binom{5}{3} = 200, and (62)(54)=75\binom{6}{2}\binom{5}{4} = 75, for a total of 150+200+75=425150 + 200 + 75 = 425 boards. Next, we determine how many of these boards contain both senior analysts A and B. Fixing A and B requires choosing 4 more members from the remaining 4 seniors and 5 juniors such that the total junior count is at least 2. The valid remaining senior choices s′s' are 0, 1, or 2, yielding (40)(54)=5\binom{4}{0}\binom{5}{4} = 5, (41)(53)=40\binom{4}{1}\binom{5}{3} = 40, and (42)(52)=60\binom{4}{2}\binom{5}{2} = 60, totaling 5+40+60=1055 + 40 + 60 = 105 boards. Subtracting these forbidden boards from 425 yields 425−105=320425 - 105 = 320.

Step-by-Step Solution

1
Calculate total valid boards with at least 2 senior analysts and at least 2 junior analysts without restriction on A and B.
Total valid boards without conflict constraint = 425
The valid (Senior, Junior) distributions for a 6-member board are (4,2), (3,3), and (2,4). Their combinations are (64)(52)=150\binom{6}{4}\binom{5}{2} = 150, (63)(53)=200\binom{6}{3}\binom{5}{3} = 200, and (62)(54)=75\binom{6}{2}\binom{5}{4} = 75, giving 150+200+75=425150 + 200 + 75 = 425.
2
Calculate the number of valid boards that contain BOTH senior analysts A and B.
Number of invalid boards containing both A and B = 105
If A and B are both selected, 2 senior spots are fixed. We need 4 remaining members from the remaining 4 seniors and 5 juniors. To satisfy the junior constraint (at least 2 juniors), the additional senior count s′s' can be 0, 1, or 2: s′=0  ⟹  (40)(54)=5s'=0 \implies \binom{4}{0}\binom{5}{4}=5, s′=1  ⟹  (41)(53)=40s'=1 \implies \binom{4}{1}\binom{5}{3}=40, s′=2  ⟹  (42)(52)=60s'=2 \implies \binom{4}{2}\binom{5}{2}=60. Summing gives 5+40+60=1055 + 40 + 60 = 105.
3
Subtract the conflicting boards containing both A and B from the total valid boards.
Valid boards = 425 - 105 = 320
Complementary counting gives the exact number of boards satisfying all conditions.

Key Concept

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

Permutations and Combinations

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