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Difficulty: HardPermutations, Combinations, and Fundamental Counting Principle

A university research department consists of 55 senior professors and 66 junior researchers. A project committee of 55 members is to be formed from this group. The committee must include at least 22 senior professors and at least 22 junior researchers. Additionally, two specific junior researchers, Alex and Blair, refuse to serve on the committee together. How many different 5-member committees can be formed under these conditions?

  1. A
    240
  2. B
    260
  3. 300Answer
  4. D
    350
  5. E
    455

Answer

300
The total number of committees satisfying the role composition rules without restrictions is 350 (200 committees with 2 seniors and 3 juniors, plus 150 committees with 3 seniors and 2 juniors). Subtracting the 50 committees that contain both Alex and Blair leaves 300 valid committees.

Step-by-Step Solution

1
Determine valid committee compositions based on role count constraints.
Two valid distributions of 5 members: Case 1 has 2 senior professors and 3 junior researchers; Case 2 has 3 senior professors and 2 junior researchers.
The committee must contain at least 2 seniors and at least 2 juniors out of 5 total members.
2
Calculate total valid committees without the adjacency/conflict restriction.
Case 1: (52)×(63)=10×20=200\binom{5}{2} \times \binom{6}{3} = 10 \times 20 = 200. Case 2: (53)×(62)=10×15=150\binom{5}{3} \times \binom{6}{2} = 10 \times 15 = 150. Total without restriction = 200+150=350200 + 150 = 350.
Using combination formula (nk)=n!k!(n−k)!\binom{n}{k} = \frac{n!}{k!(n-k)!} to count valid group selections.
3
Calculate the number of prohibited committees containing both Alex and Blair.
For Case 1 (2 seniors, 3 juniors): choose 2 seniors from 5 and 1 additional junior from the remaining 4, giving (52)×(41)=10×4=40\binom{5}{2} \times \binom{4}{1} = 10 \times 4 = 40. For Case 2 (3 seniors, 2 juniors): choose 3 seniors from 5 and 0 additional juniors from the remaining 4, giving (53)×(40)=10×1=10\binom{5}{3} \times \binom{4}{0} = 10 \times 1 = 10. Total restricted committees = 40+10=5040 + 10 = 50.
When Alex and Blair are both selected, 2 junior slots are fixed, leaving remaining slots to be filled from the remaining 4 junior researchers.
4
Subtract restricted committees from total valid composition committees.
350−50=300350 - 50 = 300.
Complementary counting yields the total number of valid committees satisfying all constraints.

Key Concept

Combinations with Composition and Exclusion Restrictions
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Topics

Permutations, Combinations, and Fundamental Counting Principle

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