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\mathrm{m} men and \mathrm{w} women are to be seated in a row so that no two women sit together. If \mathrm{m>w}, then the number of ways in which they can be seated is

Option: 1

\mathrm{\frac{m !(m+1) !}{(m-w+1) !}}


Option: 2

\mathrm{{ }^{m} C_{m-w}(m-w) !}


Option: 3

\mathrm{{ }^{m+w} C_{m}(m-w) !}


Option: 4

none of these


Answers (1)

best_answer

We first arrange the \mathrm{m} men.
This can be done in \mathrm{m!} ways. After \mathrm{m} men have taken their seats, the women must choose \mathrm{w} seats out of \mathrm{(m+1)}\mathrm{(m+1)}  seats marked with \mathrm{X} below.

\mathrm{X M X M X M X \ldots X M X}
1st  2nd    3rd             \mathrm{m} th

They can choose \mathrm{w} seats in \mathrm{{ }^{m+1} C_{w}}, ways and take their seats in \mathrm{w!} ways.
Thus, the required number of arrangements is

\mathrm{m !\left({ }^{m+1} C_{w}\right)(w !)=\frac{m !(m+1) ! w !}{w !(m+1-w) !}=\frac{m !(m+1) !}{(m+1-w) !}}

Posted by

Ritika Kankaria

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