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

Question

If $\mathrm{m}$ men and $\mathrm{n}$ women are to be seated in a row so that no two women sit together. If $m > n$, then the number of ways in which they can be seated is, $\frac{\mathrm{m} ! \mathrm{n} !}{(\mathrm{m}-\mathrm{n}+1) !}$, $\frac{(\mathrm{m}+1) !(\mathrm{n}+1) !}{(\mathrm{m}-\mathrm{n}+1) !}$, $\frac{\mathrm{m} !(\mathrm{m}+1) !}{(\mathrm{m}-\mathrm{n}+1) !}$, None of these

MARKS App · Accepted Answer

The $\mathrm{m}$ men can be seated in $\mathrm{m}$ ! ways When they are seated, there are $(\mathrm{m}+1)$ places, shown by $\times$ where $\mathrm{n}$ women can sit. Then no two women would be together as shown below $\times \mathrm{M} \times \mathrm{M} \times \mathrm{M} \times \ldots \ldots \times \mathrm{M} \times$ Then $\mathrm{n}$ women can arrange themselves in these $(\mathrm{m}+1)$ places in ${ }^{\mathrm{m}+1} \mathrm{P}_{\mathrm{n}}$ ways. Hence the required no. of ways $=m ! \times{ }^{m+1} P_n=\frac{m !(m+1) !}{(m-n+1) !}$

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