There are three men and seven women taking a dance class. Number of different ways in which each man is paired with a woman partner, and the four remaining women are paired into two pairs each of two is

Options

A. A. 105
B. B. 360
C. C. 630
D. D. 512

Answer

To solve the problem of pairing three men with three women and then pairing the remaining four women into two pairs, we can break it down into a series of steps.

### Step-by-Step Solution:

1. **Select 3 Women from 7:**
We need to select 3 women from the 7 available women. The number of ways to choose 3 women from 7 is given by the combination formula:
\[ \text{Number of ways} = \binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!} = 35 \]

**Hint:** Use the combination formula \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\) to find the number of ways to choose a subset from a larger set.

2. **Pair the Selected Women with the Men:**
After selecting 3 women, we need to pair them with the 3 men. The number of ways to pair 3 men with 3 women is given by the number of permutations of the 3 women:
\[ \text{Number of ways} = 3! = 6 \]

**Hint:** The number of ways to arrange \(n\) items is \(n!\). For pairing, think of it as arranging the selected women.

3. **Pair the Remaining 4 Women:**
Now, we have 4 women left, and we need to pair them into 2 pairs. To find the number of ways to pair 4 women, we can use the following approach:
- First, arrange the 4 women, which can be done in \(4!\) ways.
- Since we are forming pairs, we need to divide by the number of ways to arrange the pairs (which is \(2!\)) and also by the number of ways to arrange the 2 pairs themselves (which is also \(2!\)):
\[ \text{Number of ways} = \frac{4!}{2! \cdot 2!} = \frac{24}{4} = 6 \]

**Hint:** When forming pairs, remember to account for the indistinguishability of the pairs by dividing by the factorial of the number of pairs.

4. **Calculate the Total Number of Ways:**
Now, we can calculate the total number of ways to arrange everything:
\[ \text{Total ways} = \binom{7}{3} \times 3! \times \frac{4!}{2! \cdot 2!} = 35 \times 6 \times 6 = 1260 \]

**Hint:** Multiply the number of ways from each step to get the total number of arrangements.

### Final Answer:
The total number of different ways in which each man is paired with a woman partner and the four remaining women are paired into two pairs is **1260**.