Theory question 109
Permutations and CombinationsA committee of 6 people is to be chosen from 7 mid-fielders, 5 attackers and 8 referees. If each discipline must be represented, determine the number of ways of selecting the committee to include: (a) exactly one attacker and one mid-fielder (b) exactly two referees (c) at least two persons from the attackers
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(a) 1 mid-fielder, 1 attacker, 4 referees: 7 × 5 × ⁸C₄ = 7 × 5 × 70 = 2450.
(b) 2 referees (⁸C₂ = 28) and 4 from mid-fielders and attackers with at least one of each: ¹²C₄ − ⁷C₄ − ⁵C₄ = 495 − 35 − 5 = 455. Total 28 × 455 = 12 740.
(c) 2 attackers (⁵C₂ = 10): (7×56 + 21×28 + 35×8) = 1260, giving 12 600;
3 attackers (10): (7×28 + 21×8) = 364, giving 3 640;
4 attackers (5): 7×8 = 56, giving 280;
5 attackers leaves one place, so not possible.
Total = 12 600 + 3 640 + 280 = 16 520.
Every committee needs at least one of each group, so list the allowed splits.