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:
at most $3$ 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:
at most $3$ girls?
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 $^{4} C_{3} \times^{9} C_{4}$ ways.
$2$ girls and $5$ boys can be selected in $^{4} C_{2} \times^{9} C_{5}$ ways.
$1$ girl and $6$ boys can be selected in $^{4} C_{1} \times^{9} C_{6}$ ways.
No girl and $7$ boys can be selected in $^{4} C_{0} \times^{9} C_{7}$ ways.
Therefore, in this case, required number of ways
$=^{4} C_{3} \times^{9} C_{4}+^{4} C_{2} \times^{9} C_{5}+^{4} C_{1} \times^{9} C_{6}+^{4} C_{0} \times^{9} C_{7}$
$=\frac{4 !}{3 ! 1 !} \times \frac{9 !}{4 ! 5 !}+\frac{4 !}{2 ! 2 !} \times \frac{9 !}{5 ! 4 !}+\frac{4 !}{1 ! 3 !} \times \frac{9 !}{6 ! 3 !}+\frac{4 !}{0 ! 4 !} \times \frac{9 !}{7 ! 2 !}$
$=504+756+336+36$
$=1632$
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