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

Solution:

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

1. 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 $={ }^{4} \mathrm{C}_{3} \times{ }^{9} \mathrm{C}_{4}=\frac{4 !}{3 ! 1 !} \times \frac{9 !}{4 ! 5 !}$

$=4 \times \frac{9 \times 8 \times 7 \times 6 \times 5 !}{4 \times 3 \times 2 \times 1 \times 5 !}$

$=504$

(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 boy

(c) 1 girl and 6 boys

(d) No girl and 7 boys

3 girls and 4 boys can be selected in ${ }^{4} \mathrm{C}_{3} \times{ }^{9} \mathrm{C}_{4}$ ways.

2 girls and 5 boys can be selected in ${ }^{4} \mathrm{C}_{2} \times{ }^{9} \mathrm{C}_{5}$ ways.

1 girl and 6 boys can be selected in ${ }^{4} \mathrm{C}_{1} \times{ }^{9} \mathrm{C}_{6}$ ways.

No girl and 7 boys can be selected in ${ }^{4} \mathrm{C}_{0} \times{ }^{9} \mathrm{C}_{7}$ ways.

Therefore, in this case, required number of ways

$={ }^{4} \mathrm{C}_{3} \times{ }^{9} \mathrm{C}_{4}+{ }^{4} \mathrm{C}_{2} \times{ }^{9} \mathrm{C}_{5}+{ }^{4} \mathrm{C}_{1} \times{ }^{9} \mathrm{C}_{6}+{ }^{4} \mathrm{C}_{0} \times{ }^{9} \mathrm{C}_{7}$

$=\frac{4 !}{3 ! 1 !} \times \frac{9 !}{4 ! 5 !}+\frac{4 !}{2 ! 2 !} \times \frac{9 !}{5 ! 4 !}+\frac{4 !}{1 ! 3 !} \times \frac{9 !}{6 ! 3 !}+\frac{4 !}{0 ! 4 !} \times \frac{9 !}{7 ! 2 !}$

$=504+756+336+36$

 

$=1632$