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A group consists of $4$ girls and $7$ boys. In how many ways can a team of $5$ members be selected if the team has at least one boy and one girl ?
$441$
$441$
$441$
$441$
Solution
since, at least one boy and one girl are to be there in every team. Therefore, the team can consist of
$(a)$ $1$ boy and $4$ girls
$(b)$ $2$ boys and $3$ girls
$(c)$ $3$ boys and $2$ girls
$(d)$ $4$ boys and $1$ girl.
$1$ boy and $4$ girls can be selected in $^{7} C _{1} \times^{4} C _{4}$ ways.
$2$ boys and $3$ girls can be selected in $^{7} C _{2} \times^{4} C _{3}$ ways.
$3$ boys and $2$ girls can be selected in $^{7} C _{3} \times^{4} C _{2}$ ways.
$4$ boys and $1$ girl can be selected in $^{7} C _{4} \times^{4} C _{1}$ ways.
Therefore, the required number of ways
$=\,^{7} C _{1} \times^{4} C _{4}+^{7} C _{2} \times^{4} C _{3}+^{7} C _{3} \times^{4} C _{2}+^{7} C _{4} \times^{4} C _{1}$
$=7+84+210+140=441$