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Question
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 ______.
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Solution
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 7800.
Explanation:
We have 10 men and 7 women out of which a committee of 6 is to be formed which contain atleast 3 men and 2 women
Therefore, Number of ways = 10C3 × 7C3 + 10C4 × 7C2
= `(10 xx 9 xx 8)/(3 xx 2 xx 1) xx (7 xx 6 xx 5)/(3 xx 2 xx 1) + (10 xx 9 xx 8 xx 7)/(4 xx 3 xx 2 xx ) xx (7 xx 6)/(2 xx 1)`
= 120 × 35 + 210 × 21
= 4200 + 4410
= 8610
If 2 particular women to be always present, then the number of ways = 10C4 × 5C0 + 10C3 × 5C1
= `(10 xx 9 xx 8 xx 7)/(4 xx 3 xx 2 xx 1) xx 1 + (10 xx 9 xx 8)/(3 xx 2 xx 1) xx 5`
= 210 + 120 × 5
= 210 + 600
= 810
∴ Total number of committee = 8610 – 810 = 7800
Hence, the value of the filler is 7800.
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