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Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
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The number of ways of dividing 15 men and 15 women into 15 couples each consisting of a man and woman is

$\begin{array}{1 1}(A)\;1240\\(B)\;1840\\(C)\;1820\\(D)\;2005\end{array} $

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  • $C(n,r)=\large\frac{n!}{r!(n-r)!}$
The number of ways of choosing first couple is =$15C_1\times 15C_1$
$\Rightarrow (15)^2$
The number of ways of choosing second couple is =$14C_1\times 14C_1$
$\Rightarrow (14)^2$
Thus the number of ways of choosing the couple is =$15^2+14^2+13^2+.....+2^2+1^2$
$\Rightarrow \large\frac{15\times(15+1)[2(15+1)]}{6}$
$\Rightarrow 1240$
Hence (A) is the correct answer.
answered May 19, 2014 by sreemathi.v

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