The number of ways in which 15 boys and 2 girls can sit in a row such that between the girls at the most 2 boys sit is

Options

A. `17!-(12!xx3!)`
B. `17!-(.^(12)C_()xx3!)`
C. `17!-(.^(12)C_(3)xx15!)`
D. `17!-(9!xx2!xx15!)`

Answer

To solve the problem of how many ways 15 boys and 2 girls can sit in a row such that at most 2 boys sit between the girls, we can break it down into three cases based on the number of boys sitting between the girls.

### Step-by-Step Solution

**Step 1: Case 1 - No Boys Between the Girls**

In this case, the two girls sit next to each other. We can treat the two girls as a single unit or block. Therefore, we have:

- 15 boys + 1 block of girls = 16 units to arrange.

The number of ways to arrange these 16 units is \(16!\). Additionally, the two girls can switch places, which gives us an additional factor of \(2!\).

Thus, the total for Case 1 is:
\[ \text{Ways for Case 1} = 16! \times 2! \]

**Step 2: Case 2 - One Boy Between the Girls**

In this case, we have one boy sitting between the two girls. We can represent this arrangement as:

- Girl - Boy - Girl

Here, we first choose 1 boy from the 15 boys. The number of ways to choose 1 boy from 15 is \( \binom{15}{1} \).

Now we have:
- 14 remaining boys + 1 block (Girl-Boy-Girl) = 15 units to arrange.

The number of ways to arrange these 15 units is \(15!\). The two girls can also switch places, giving us another factor of \(2!\).

Thus, the total for Case 2 is:
\[ \text{Ways for Case 2} = \binom{15}{1} \times 15! \times 2! \]

**Step 3: Case 3 - Two Boys Between the Girls**

In this case, we have two boys sitting between the two girls. We can represent this arrangement as:

- Girl - Boy - Boy - Girl

We first choose 2 boys from the 15 boys. The number of ways to choose 2 boys from 15 is \( \binom{15}{2} \).

Now we have:
- 13 remaining boys + 1 block (Girl-Boy-Boy-Girl) = 14 units to arrange.

The number of ways to arrange these 14 units is \(14!\). The two girls can also switch places, giving us another factor of \(2!\).

Thus, the total for Case 3 is:
\[ \text{Ways for Case 3} = \binom{15}{2} \times 14! \times 2! \]

**Step 4: Total Number of Ways**

Now, we sum the total number of ways from all three cases:
\[ \text{Total Ways} = \text{Ways for Case 1} + \text{Ways for Case 2} + \text{Ways for Case 3} \]

Substituting the expressions we found:
\[ \text{Total Ways} = 16! \times 2! + \binom{15}{1} \times 15! \times 2! + \binom{15}{2} \times 14! \times 2! \]

**Step 5: Simplifying the Expression**

Calculating each term:
1. \(16! \times 2! = 16! \times 2\)
2. \(\binom{15}{1} = 15\), so \(15 \times 15! \times 2! = 15 \times 15! \times 2\)
3. \(\binom{15}{2} = \frac{15 \times 14}{2} = 105\), so \(105 \times 14! \times 2!\)

Putting it all together:
\[ \text{Total Ways} = 16! \times 2 + 15 \times 15! \times 2 + 105 \times 14! \times 2 \]

Factoring out \(2\):
\[ \text{Total Ways} = 2 \times (16! + 15 \times 15! + 105 \times 14!) \]

### Final Answer

The total number of ways in which 15 boys and 2 girls can sit in a row such that at most 2 boys sit between the girls is given by the expression above.