Quantity I: Number of ways to arrange 5 men and 5 women so that no two men or two women sit next to each other.

Quantity II: Number of ways to arrange 5 men and 5 women such that all men sit together.

Which of the following is true?

Question

Quantity I: Number of ways to arrange 5 men and 5 women so that no two men or two women sit next to each other.

Quantity II: Number of ways to arrange 5 men and 5 women such that all men sit together.

Which of the following is true?

Accepted Answer

**Solution:**

1. **Quantity I:** Number of ways to arrange 5 men and 5 women so that no two men or two women are adjacent.

- First, arrange the 5 men in a row: number of ways = $5!$
- To ensure no two women sit together, women must be placed in the gaps between men. There are 6 such gaps (before, between, and after men).
- Since there are 5 women, they can be placed in these 6 gaps such that no two women are adjacent only if women occupy alternate positions with men.
- The only way to do this is to arrange men and women alternately, starting either with a man or a woman.
- Number of ways to arrange women = $5!$
- Since men can be placed first or women can be placed first, multiply by 2.

Therefore, total ways for Quantity I = $2 	imes 5! 	imes 5!$

2. **Quantity II:** Number of ways to arrange 5 men and 5 women such that all men sit together.

- Treat the 5 men as a single block. Along with 5 women, total entities = 6 (1 block + 5 women).
- Number of ways to arrange these 6 entities = $6!$
- Inside the men block, men can be arranged among themselves in $5!$ ways.

Therefore, total ways for Quantity II = $6! 	imes 5!$

3. **Comparison:**

- Calculate factorial values:
  - $5! = 120$
  - $6! = 720$

- Quantity I = $2 	imes 120 	imes 120 = 2 	imes 14400 = 28800$
- Quantity II = $720 	imes 120 = 86400$

Since $28800 < 86400$, Quantity I < Quantity II.

**Answer:** (b) Quantity I < Quantity II

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