If $p > q$ , the number of ways of $p$ men and $q$ women can be seated in a row so that no two women sit together is

\frac{p! q!}{(p + q)!}

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\frac{(p + q)!}{p! (q +!)!}

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\frac{p! (q + 1)!}{(p - q + 1)!}

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\frac{p! (p + 1)!}{(p - q + 1)!}

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Solution

The correct option is C $\frac{p! (p + 1)!}{(p - q + 1)!}$
Arrangement of $p$ men in a row $= p!$
$∴$ Required number of ways $= {p!}^{p + 1} P_{q} = \frac{p! (p + 1)!}{(p - q + 1)!}$

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If p>q, the number of ways of p men and q women can be seated in a row so that no two women sit together is

The correct option is C p!(p+1)!(p−q+1)!Arrangement of p men in a row =p!∴ Required number of ways =p!p+1Pq=p!(p+1)!(p−q+1)!

If $p > q$ , the number of ways of $p$ men and $q$ women can be seated in a row so that no two women sit together is

The correct option is C $\frac{p! (p + 1)!}{(p - q + 1)!}$
Arrangement of $p$ men in a row $= p!$
$∴$ Required number of ways $= {p!}^{p + 1} P_{q} = \frac{p! (p + 1)!}{(p - q + 1)!}$