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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 one boy and one girl

$\begin{array}{1 1}(A)\;430\\(B)\;440\\(C)\;441\\(D)\;550\end{array} $

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  • $C(n,r)=\large\frac{n!}{r!(n-r)!}$
Since at least one boy and one girl are to be there in every team
$\therefore$ The team consist of
(a) 1 boy 4 girls $\rightarrow 7C_1\times 4C_4$
(b) 2 boys and 3 girls $\rightarrow 7C_2\times 4C_3$
(c) 3 boys and 2 girls $\rightarrow 7C_3\times 4C_2$
(d) 4 boys and 1 girl $\rightarrow 7C_4\times 4C_1$
$\therefore$ The required number of ways =$7C_1\times 4C_4+7C_2\times 4C_3+7C_3\times 4C_2+7C_4\times 4C_1$
$7C_1=\large\frac{7!}{1!6!}$$=7$
$4C_4=\large\frac{4!}{4!0!}$$=1$
$7C_2=\large\frac{7!}{2!5!}$$=21$
$4C_3=\large\frac{4!}{3!1!}=\frac{4\times 3!}{3!\times 1}$$=4$
$7C_3=\large\frac{7!}{3!4!}=\frac{7\times 6\times 5\times 4!}{3\times 2\times 4!}$$=35$
$4C_2=\large\frac{4!}{2!2!}=\frac{4\times 3\times 2!}{2\times 1\times 2!}$$=6$
$7C_4=\large\frac{7!}{4!3!}=\frac{7\times 6\times 5\times 4!}{4!\times 3\times 2}$$=35$
$4C_1=\large\frac{4!}{1!3!}=\frac{4\times 3!}{1!\times 3!}$$=4$
$\Rightarrow 7\times 1+21\times 4+35\times 6+35\times 4$
$\Rightarrow 7+84+210+140$
$\Rightarrow 441$ ways
Hence (C) is the correct answer.
answered May 16, 2014 by sreemathi.v
 

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