(i) Given exactly $3$ girls

Total numbers of girls are $4$

Out of which $3$ are to be chosen

∴ Number of ways in which choice would be made

\[\Rightarrow {{~}^{4}}{{C}_{3}}=\]

Numbers of boys are $9$ out of which $4$ are to be chosen which is given by

\[^{9}{{C}_{4}}\]

Total ways of forming the committee with exactly three girls

= ⁴C₃ × ⁹C₄

= 

(ii) Given at least $3$ girls

There are two possibilities of making committee choosing at least $3$ girls

There are $3$ girls and $4$ boys or there are $4$ girls and $3$ boys

Choosing three girls we have done in (i)

Choosing four girls and $3$ boys would be done in

\[^{4}{{C}_{4}}\]

 ways

And choosing $3$ boys would be done in

\[^{9}{{C}_{3}}\]

Total ways =

\[^{4}{{C}_{4}}~\times {{~}^{9}}{{C}_{3}}\]

Total numbers of ways of making the committee are

\[504\text{ }+\text{ }84\text{ }=\text{ }588\]