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Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
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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 three girls

$\begin{array}{1 1}(A)\;90\\(B)\;91\\(C)\;92\\(D)\;93\end{array} $

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  • $C(n,r)=\large\frac{n!}{r!(n-r)!}$
Since the team has to consist of at least three girls
The team can consist of
(a) 3 girls and 2 boys
(b) 4 girls and 1 boy
Note : The team cannot have all 5 girls,because the group has only 4 girls.
3 girls and 2 boys can be selected in =$4C_3\times 7C_2$ ways
4 girls and 1 boy can be selected in =$4C_4\times 7C_1$ ways
$\therefore$ The required number of ways =$4C_3\times 7C_2+4C_4\times 7C_1$
$4C_3=\large\frac{4!}{3!1!}=\frac{4\times 3!}{3!}$$=4$
$7C_2=\large\frac{7!}{2!5!}=\frac{7\times 6\times 5!}{2\times 5!}$$=21$
$4C_4=\large\frac{4!}{4!0!}$$=1$
$7C_1=\large\frac{7!}{1!6!}=\frac{7\times 6!}{6!}$$=7$
$\Rightarrow 4\times 21+1\times 7$
$\Rightarrow 84+7$
$\Rightarrow 91$ ways
Hence (B) is the correct answer.
answered May 16, 2014 by sreemathi.v
 

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