In a party of New Year's Eve, if men shook hands among themselves , there would be 21 handshakes in all . However , if the men shook hands with the women, there would be 35 handshakes . How many handshakes would have happened if the women shook hands among themselves ?

Options

A. 5
B. 7
C. 10
D. 12

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Understand the Problem
We need to find out how many handshakes would occur if the women shook hands among themselves, given the number of handshakes when men shook hands among themselves and when men shook hands with women.

### Step 2: Set Up Variables
Let:
- \( M \) = number of men
- \( W \) = number of women

### Step 3: Handshakes Among Men
The number of handshakes among men is given by the combination formula:
\[ H(M, M) = \frac{M(M-1)}{2} = 21 \]
This means:
\[ M(M-1) = 42 \quad \text{(Multiplying both sides by 2)} \]

### Step 4: Solve for M
We need to solve the equation:
\[ M^2 - M - 42 = 0 \]
We can factor this quadratic equation:
\[ (M - 7)(M + 6) = 0 \]
Thus, \( M = 7 \) or \( M = -6 \). Since the number of men cannot be negative, we have:
\[ M = 7 \]

### Step 5: Handshakes Between Men and Women
Now, we use the information about handshakes between men and women:
\[ H(M, W) = M \times W = 35 \]
Substituting \( M = 7 \):
\[ 7 \times W = 35 \]
Solving for \( W \):
\[ W = \frac{35}{7} = 5 \]

### Step 6: Handshakes Among Women
Now we need to find the number of handshakes among women:
\[ H(W, W) = \frac{W(W-1)}{2} \]
Substituting \( W = 5 \):
\[ H(5, 5) = \frac{5(5-1)}{2} = \frac{5 \times 4}{2} = \frac{20}{2} = 10 \]

### Final Answer
The number of handshakes that would have happened if the women shook hands among themselves is:
\[ \boxed{10} \]

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