The number of ways in which 10 boys can be divied into 2 groups of 5, such that two tallest boys are in two different groups, is equal to

Options

A. 70
B. 35
C. 252
D. 126

Answer

To solve the problem of dividing 10 boys into 2 groups of 5 such that the two tallest boys are in different groups, we can follow these steps:

### Step 1: Exclude the Two Tallest Boys
First, we exclude the two tallest boys from the group. This leaves us with 8 boys.

### Step 2: Divide the Remaining Boys
Next, we need to divide these 8 boys into two groups of 4 each. The number of ways to choose 4 boys from 8 is given by the combination formula:

\[ \text{Number of ways} = \binom{8}{4} \]

### Step 3: Calculate the Combination
The combination \(\binom{8}{4}\) can be calculated as:

\[ \binom{8}{4} = \frac{8!}{4! \cdot 4!} \]

### Step 4: Account for the Two Tallest Boys
Now, we need to include the two tallest boys (let's call them B and C). Since they must be in different groups, we can place one tallest boy in one group and the other tallest boy in the other group. There are 2 ways to do this (B in Group 1 and C in Group 2, or B in Group 2 and C in Group 1).

### Step 5: Combine the Results
Now, we combine the results from the previous steps. The total number of ways to divide the boys is given by:

\[ \text{Total Ways} = \binom{8}{4} \times 2 \]

### Step 6: Calculate the Total Ways
Now we can calculate the total number of ways:

1. Calculate \(\binom{8}{4}\):

\[ \binom{8}{4} = \frac{8!}{4! \cdot 4!} = \frac{40320}{24 \cdot 24} = \frac{40320}{576} = 70 \]

2. Multiply by 2 (for the two tallest boys):

\[ \text{Total Ways} = 70 \times 2 = 140 \]

### Final Answer
Thus, the total number of ways in which the 10 boys can be divided into 2 groups of 5, such that the two tallest boys are in different groups, is **140**.

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