Question

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:
(i) exactly 3 girls?
(ii) atleast 3 girls?
(iii) atmost 3 girls?

Answer 1

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

**(i) ** Since exactly 3 girls are to be there in every committee, each committee must consist of (7 - 3) = 4 boys only.
Thus, in this case, required number of ways = $^4C_3\times\space^9C_4=\frac{4!}{3!1!}\times\frac{9!}{4!5!}$
$=4\times\frac{9\times8\times7\times5!}{4\times3\times2\times1\times5!}$
$=504$

**(ii) ** Since at least 3 girls are to be there in every committee, the committee can consist of
(a) 3 girls and 4 boys or
(b) 4 girls and 3 boys
3 girls and 4 boys can be selected in $^4C_3\times\space^9C_4$ ways.
4 girls and 3 boys can be selected in $^4C_4\times \space^9C_3$ ways.
Therefore, in this case, required number of ways = $^4C_3\times\space^9C_4+^4C_4\times\space^9C_3$
$=504+84=588$

(iii) Since atmost 3 girls are to be there in every committee, the committee can consist of
(a) 3 girls and 4 boys
(b) 2 girls and 5 boys
(c) 1 girl and 6 boys
(d) No girl and 7 boys 3 girls and 4 boys can be selected in $^4C_3\times\space^9C_4$ ways.
2 girls and 5 boys can be selected in $^4C_2\times\space^9C_5$ ways.
1 girl and 6 boys can be selected in $^4C_1\times\space^9C_6$ ways.
No girl and 7 boys can be selected in $^4C_0\times\space^9C_7$ ways.
Therefore, in this case, required number of ways

=$$^4C_3\times\space^9C_4+^4C_2\times^9C_5+^4C_1\times\space^9C_6+^4C_0\times\space^9C_7
=$$\frac{4!}{3!1!}\times\frac{9!}{4!5!}$$+\frac{4!}{2!2!}\times\frac{9!}{5!4!}$$+\frac{4!}{1!3!}\times\frac{9!}{6!3!}+\frac{4!}{0!4!}\times\frac{9!}{7!2!}
$=504+756+336+36$
$=1632$