m men and n women are to be seated in a...

MathematicsLinear PermutationsJEE Advanced 1983Easy

$m$ men and $n$ women are to be seated in a row so that no two women sit together. If $m > n$ , then show that the number of ways in which they can be seated is $\frac{m ! ( m + 1 )!}{( m - n + 1 )!}$

Visualized Solution (English)

Visualizing the Arrangement

Total Men: $m$
Total Women: $n$
Constraint: No two women sit together.
Strategy: Use the Gap Method by seating men first.

Arranging .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m Men

Number of ways to arrange $m$ men in a row: $m!$

Identifying the Gaps

Number of gaps created by $m$ men: $m + 1$
These include gaps between men and the two ends of the row.

Arranging .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n Women

Number of ways to seat $n$ women in $m + 1$ gaps: $^{m + 1} P_{n}$
Since $m > n$ , $m + 1 > n$ , making the arrangement possible.

Final Calculation

Total ways $= m! \times^{m + 1} P_{n}$
Using $^{n} P_{r} = \frac{n !}{( n - r )!}$ :
Total ways $= m! \times \frac{( m + 1 )!}{( m + 1 - n )!}$
Final Expression: $\frac{m ! ( m + 1 )!}{( m - n + 1 )!}$

Key Takeaways

Gap Method Rule: Arrange unconstrained items first, then place constrained items in gaps.
Linear vs Circular: In a row, $m$ items create $m + 1$ gaps. In a circle, $m$ items create only $m$ gaps.
Next Challenge: How would the answer change if the women were identical?

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