Question:hard

The number of ways in which a committee of 3 ladies and 4 gentlemen can be appointed from a meeting consisting of 8 ladies and 7 gentlemen, if Mrs. X refuses to serve in a committee if Mr. Y is its member, is

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Split into two cases, Mr. Y included or excluded, since Mrs. X's availability depends entirely on whether Mr. Y is on the committee.
Updated On: Jul 16, 2026
  • 1960
  • 3240
  • 1540
  • none of these
Show Solution

The Correct Option is C

Solution and Explanation

A cleaner way to handle this restriction is complementary counting: find the total number of committees with no restriction, then subtract the invalid committees where both Mrs. X and Mr. Y are members together.

  1. Total unrestricted committees: choose 3 ladies from 8 and 4 gentlemen from 7, giving \( {}^8C_3 \times {}^7C_4 = 56 \times 35 = 1960\).
  2. Invalid committees (X and Y both included): fix Mrs. X as one of the 3 ladies, so the other 2 ladies come from the remaining 7, giving \( {}^7C_2 = 21\) ways. Fix Mr. Y as one of the 4 gentlemen, so the other 3 gentlemen come from the remaining 6, giving \( {}^6C_3 = 20\) ways. Invalid committees = 21 x 20 = 420.

Valid committees = 1960 - 420 = 1540, which exactly matches the case-by-case count and confirms option C.

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