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Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
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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 most $3$ girls?(i) No girl,$7$ boys, (ii) $1$ girl,$6$ boys,(iii) $2$ girls,$5$ boys,(iv) $3$ girls,$4$ boys

$\begin{array}{1 1}(A)\;1630\\(B)\;1631\\(C)\;1633\\(D)\;1632\end{array} $

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1 Answer

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This can be done in $9C_7+4C_1\times 9C_6+4C_2\times 9C_5+4C_3 \times 9C_4$
$\Rightarrow 9C_2+4C_1\times 9C_3+4C_2 \times 9C_4 +4C_1\times 9C_4$
$\Rightarrow \large\frac{9\times 8}{1\times 2}+\frac{4}{1}+\frac{9\times 8\times 7}{1\times 2\times 3 \times 4}+\frac{4}{1}\times \frac{9\times 8\times 7\times 6}{1\times 2\times 3\times 4}$
$\Rightarrow 36+4\times 84 +6\times 126+4\times 126$ ways
$\Rightarrow 36+336+126(6+4)$ ways
$\Rightarrow 36+336+1260$
$\Rightarrow 1632$ ways
Hence (D) is the correct answer.
answered May 14, 2014 by sreemathi.v
 

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