Five men and five women are be arranged in a row while seating in a party.
Quantity I: number of ways of arranging 5 men and 5 women such that no two men or women are adjacent to each other.
Quantity II: Number of ways of arranging 5 men and 5 women such that all men sit together.

Options

A. Quantity I `gt` Quantity II
B. Quantity I `lt` Quantity II
C. Quantity I `ge` Quantity II
D. Quantity I `le` Quantity II

Explanation

To solve the problem, we need to calculate the two quantities separately.

### Step 1: Calculate Quantity I
**Quantity I** is the number of ways to arrange 5 men and 5 women such that no two men or women are adjacent to each other.

1. **Arrange the Women**: First, we can arrange the 5 women. The number of ways to arrange 5 women is given by \(5!\).
\[ 5! = 120 \]

2. **Create Spaces for Men**: Once the women are arranged, they create spaces for the men. For 5 women, there are 6 potential spaces (before the first woman, between women, and after the last woman) to place the men:
- _ W _ W _ W _ W _ W _
- This gives us 6 spaces.

3. **Choose Spaces for Men**: We need to choose 5 out of these 6 spaces to place the men. The number of ways to choose 5 spaces from 6 is given by \(\binom{6}{5}\).
\[ \binom{6}{5} = 6 \]

4. **Arrange the Men**: The number of ways to arrange the 5 men in the chosen spaces is \(5!\).
\[ 5! = 120 \]

5. **Total Arrangements for Quantity I**: Therefore, the total number of arrangements for Quantity I is:
\[ 5! \times \binom{6}{5} \times 5! = 120 \times 6 \times 120 = 86400 \]

### Step 2: Calculate Quantity II
**Quantity II** is the number of ways to arrange 5 men and 5 women such that all men sit together.

1. **Treat Men as a Single Unit**: Since all men must sit together, we can treat them as a single unit or block. This gives us 6 units to arrange: the block of men and the 5 individual women.

2. **Arrange the Units**: The number of ways to arrange these 6 units (1 block of men + 5 women) is \(6!\).
\[ 6! = 720 \]

3. **Arrange the Men Within the Block**: The men within their block can be arranged among themselves in \(5!\) ways.
\[ 5! = 120 \]

4. **Total Arrangements for Quantity II**: Therefore, the total number of arrangements for Quantity II is:
\[ 6! \times 5! = 720 \times 120 = 86400 \]

### Conclusion
Now we compare the two quantities:
- **Quantity I**: 86400
- **Quantity II**: 86400

Thus, **Quantity I = Quantity II**.

### Final Answer
Both quantities are equal.

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