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It is required to sit 5 men and 4 women in a row so that the woman occupy the even places.How many such arrangemens are possible?

$\begin{array}{1 1}(A)\;2660\\(B)\;2770\\(C)\;2880\\(D)\;2990\end{array} $

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  • $n!=n(n-1)(n-2).....(3)(2)(1)$
According to the problem women will sit at the even places (2,4,6 and 8).
They can be seated in P(4,4)=4!=24 ways
Men will sit on the odd seats =(1,3,5,7 & 9 ways)
They can sit on these seats in
$\Rightarrow 5\times 4\times 3\times 2\times 1$
$\Rightarrow 120$ ways
If total number of ways P then
$P=24\times 120$
Hence (C) is the correct answer.
answered May 14, 2014 by sreemathi.v

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