<p>\(m\) men and \(n\) women are to be seated in a row so that no two women sit together. If \(m > n\), then find the number of ways in which they can be seated.</p>
Step-by-Step Solution
Key Concept: First arrange m men in a row creating (m+1) gaps (before first, between consecutive men, after last). Then select n of these gaps and arrange n women in them. This ensures no two women are adjacent.
<p><strong>Step 1:</strong> Arrange m men in a row. This can be done in m! ways.</p><p><strong>Step 2:</strong> When m men are seated in a row, they create (m+1) gaps:</p><p>_ M₁ _ M₂ _ M₃ _ ... _ Mₘ _</p><p>There are (m+1) available positions where women can sit.</p><p><strong>Step 3:</strong> We need to choose n gaps from (m+1) available gaps for n women. This can be done in P(m+1, n) ways:</p><p>P(m+1, n) = $\dfrac{(m+1)!}{(m+1-n)!}$</p><p><strong>Step 4:</strong> The total number of arrangements is:</p><p>= (Ways to arrange m men) × (Ways to choose and arrange n women in gaps)</p><p>= m! × $\dfrac{(m+1)!}{(m+1-n)!}$</p><p>= m! × $\dfrac{(m+1)!}{(m-n+1)!}$</p><p>= $\dfrac{(m+1)!\, m!}{(m-n+1)!}$</p><p>∴ Answer: $\dfrac{(m+1)!\, m!}{(m-n+1)!}$</p>
Correct Answer: \(\dfrac{(m+1)!\, m!}{(m-n+1)!}\)