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# 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: atmost 3 girls?

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:

(iii)  atmost 3 girls?

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

(ii) atmost 3 girls, there can be 4 cases :

(a) Girls =0, so boys in committee= 7-0=7

Thus, the required number of ways $=^4C_0.^9C_7$

$=\frac{4!}{4!0!}\times \frac{9!}{2!7!}$

$=9\times 4=36$

(b) Girls =1, so boys in committee= 7-1=6

Thus, the required number of ways $=^4C_1.^9C_6$

$=\frac{4!}{3!1!}\times \frac{9!}{3!6!}$

$=336$

(c) Girls =2, so boys in committee= 7-2=5

Thus, the required number of ways $=^4C_2.^9C_5$

$=\frac{4!}{2!2!}\times \frac{9!}{4!5!}$

$=756$

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

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

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

$=504$

Hence, in this case, the number of ways = 36+336+756+504=1632

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