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Question no 21 I know the answer and solved it before but now I forgot. Anyone if know the answer please tell in this thread.

Answers (7)

Correct option is D 3 ladies out of 8 can be selected in 8C3 ways and 4 gentlemen out of 7 in 7C4 ways. Now each way of selecting 3 ladies is associated with each way of selecting 4 gentlemen. Hence, the required number of ways =8C3×7C4=56×35=1960. We now find the number of committee of 3 ladies and 4 gentlemen in which both Mrs X and Mr Y are member. In this case we can select 2 other ladies from the remaining 7 in 7C2 ways and 3 other gentlemen from the remaining 6 in 6C3 ways. ∴ The number of ways in which both Mrs X and Mr Y are always included =7C2×8C3=21×20=420 Hence the required number of committee in which Mrs X And Mr Y do not serve together= 1960-420 = 1540

Thread Starter

Virendra SainiCan you please understand in more simplicity?

I am understanding but not 100%? Means little bit different in understanding?