(i) exactly \\\\(3\\\\) girls This means you have to choose \\\\(3\\\\) girls from \\\\(4\\\\) and \\\\(4\\\\) boys from \\\\(9\\\\) to form a committee of \\\\(7\\\\). This can be done in \\\\(^4{ C }_{ 3 }\ imes ^9{ C }_{ 4 }\\\\)(ii) At least \\\\(3\\\\) girls This means girls can either be \\\\(3\\\\) or \\\\(4\\\\) So first choose \\\\(3\\\\) girls and \\\\(4\\\\) boys or \\\\(4\\\\) girls and \\\\(3\\\\) boys Total ways = \\\\(^4{ C }_{ 3 }\ imes ^9{ C }_{ 4 }\\\\) + \\\\(^4{ C }_{ 4 }\ imes ^9{ C }_{ 3 }\\\\)(iii) At most \\\\(3\\\\) girls This means we have to choose \\\\(1\\\\) or \\\\(2\\\\) or \\\\(3\\\\) girls So this can be done in \\\\(^4C_0\ imes ^9C_7+^4{ C }_{ 1 }\ imes ^9{ C }_{ 6 }\\\\) + \\\\(^4{ C }_{ 2 }\ imes ^9{ C }_{ 5 }\\\\) + \\\\(^4{ C }_{ 3 }\ imes ^9{ C }_{ 4 }\\\\) ways.