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A committee of 6 is to be chosen from 10 men and 7 women so as to contain atleast 3 men and 2 women. In how many different ways can this be done if two particular women refuse to serve on the same committee.
[Hint: At least 3 men and 2 women: The number of ways = 10C3 × 7C3 + 10C4 × 7C2.
For 2 particular women to be always there: the number of ways = 10C4 + 10C3 × 5C1.
The total number of committees when two particular women are never together = Total – together.]
Formula:-
(i) nCr![]()
Number of men =10
Number of women=7
At least 3 men and 2 women: The number of ways = 10C3 × 7C3 + 10C4 × 7C2.
For 2 particular women to be always there: the number of ways = 10C4 + 10C3 × 5C1.
The total number of committees when two particular women are never together
= Total – together
=(10C3 × 7C3 + 10C4 × 7C2)-( 10C4 + 10C3 × 5C1)
=8610-810
=7800
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