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Miscellaneous Examples · Example 56

Q.There are 8 members in a committee. In how many ways we can choose:

(i) a subcommittee consisting of 3 members?
(ii) a chairperson, a secretary and a treasurer assuming that one person cannot hold more than one position?
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An unordered 3-member subcommittee is a combination (8C3{}^8C_3); filling 3 distinct named posts (order/role matters) from the same 8 members is a permutation (8P3{}^8P_3).

Combinations (order irrelevant): nCr=n!r!(n−r)!{}^nC_r=\dfrac{n!}{r!(n-r)!}. Permutations (order/role relevant, no repetition): nPr=n!(n−r)!=n×(n−1)×⋯×(n−r+1){}^nP_r=\dfrac{n!}{(n-r)!}=n\times(n-1)\times\cdots\times(n-r+1).

(i) Subcommittee of 3 members

  1. The 3 members have no distinguishing roles, so order does not matter: use combinations.
  2. 8C3=8×7×63×2×1=3366=56{}^8C_3=\dfrac{8\times7\times6}{3\times2\times1}=\dfrac{336}{6}=56.

(ii) Chairperson, secretary, treasurer (distinct posts, no repeats) …

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