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

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

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

$=504+756+336+36$

$=1632$