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?
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 $=^{4} C_{3} \times^{9} C_{4}+^{4} C_{4} \times^{9} C_{3}$
$=504+84=588$
For a scholarship, atmost $n$ candidates out of $2n+1$ can be selected. If the number of different ways of selection of atleast one candidate for scholarship is $63$, then maximum number of candidates that can be selected for the scholarship is -
The number of ways of dividing $52$ cards amongst four players equally, are
Number of integral solutions to the equation $x+y+z=21$, where $x \geq 1, y \geq 3, z \geq 4$, is equal to $..........$.
The number of values of $'r'$ satisfying $^{69}C_{3r-1} - ^{69}C_{r^2}=^{69}C_{r^2-1} - ^{69}C_{3r}$ is :-
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