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:
exactly $3$ girls $?$
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:
exactly $3$ girls $?$
A committee of $7$ has to be formed from $9$ boys and $4$ girls.
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 $=\,^{4} C_{3} \times^{9} C_{4}=\frac{4 !}{3 ! 1 !} \times \frac{9 !}{4 ! 5 !}$
$=4 \times \frac{9 \times 8 \times 7 \times 6 \times 5 !}{4 \times 3 \times 2 \times 5 !}$
$=504$
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