# Q.3.    A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways            can this be done when the committee consists of:           (ii) at least 3 girls?

S seema garhwal

There are 9 boys and 4 girls. A committee of 7 has to be formed.

(ii) at least 3 girls, there can be two cases :

(a) Girls =3, so boys in committee= 7-3=4

Thus, the required number of ways $=^4C_3.^9C_4$

$=\frac{4!}{3!1!}\times \frac{9!}{4!5!}$

$=4\times \frac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 5!}$

$= 9\times 8\times 7=504$

(b) Girls =4, so boys in committee= 7-4=3

Thus, the required number of ways $=^4C_4.^9C_3$

$=\frac{4!}{4!0!}\times \frac{9!}{3!6!}$

$= \frac{9\times 8\times 7\times 6!}{ 3\times 2\times 6!}$

$=84$

Hence, in this case, the number of ways = 504+84=588

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