Given at most $3\;girls$
In this case the numbers of possibilities are
$0\;girl\;and\;7\;boys$
$1\;girl\;and\;6\;boys$
$2\;girl\;and\;5\;boys$
$3\;girl\;and\;4\;boys$
Number of ways to choose $0\;girl\;and\;7\;boys$
\[{{=}^{4}}{{C}_{0}}~\times {{~}^{9}}{{C}_{7}}\]
Number of choosing $3\;girl\;and\;4\;boys$ has been done in (1)
\[=\text{ }504\]
Total number of ways in which committee can have at most $3\;girls$ are \[=\text{ }36\text{ }+\text{ }336\text{ }+\text{ }756\text{ }+\text{ }504\text{ }=\text{ }1632\]