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6.Permutation and Combination
medium
A committee of $7$ has to be formed from $9$ boys and $4$ girls. In how many ways can this be done when the committee consists of:
at least $3$ girls?
A
$588$
B
$588$
C
$588$
D
$588$
Solution
since at least $3$ girls are to be there in every committee, the committee can consist of
$(a)$ $3$ girls and $4$ boys or
$(b)$ $4$ girls and $3$ boys
$3$ girls and $4$ boys can be selected in $^{4} C_{3} \times^{9} C_{4}$ ways.
$4$ girls and $3$ boys can be selected in $^{4} C_{4} \times^{9} C_{3}$ ways.
Therefore, in this case, required number of ways $=^{4} C_{3} \times^{9} C_{4}+^{4} C_{4} \times^{9} C_{3}$
$=504+84=588$
Standard 11
Mathematics
