Find the number of ways in which m boys and n boys and n girls may be arranged in a row so that no two of the girls are together, it being given that `m gt n`.

Explanation

To solve the problem of arranging m boys and n girls in a row such that no two girls are together, we can follow these steps:

### Step 1: Arrange the Boys
First, we need to arrange the m boys. The number of ways to arrange m boys is given by the factorial of m, denoted as \( m! \).

**Hint:** Remember that the number of arrangements of n distinct objects is \( n! \).

### Step 2: Identify the Gaps for Girls
Once the boys are arranged, we can visualize the arrangement as creating gaps where the girls can be placed. For m boys, there will be m + 1 gaps (one before the first boy, one between each pair of boys, and one after the last boy).

**Hint:** Think about how placing boys creates spaces for girls.

### Step 3: Choose Gaps for Girls
Since we need to place n girls in these m + 1 gaps, and we want to ensure that no two girls are together, we need to choose n gaps from the available m + 1 gaps. The number of ways to choose n gaps from m + 1 is given by the combination formula \( \binom{m+1}{n} \).

**Hint:** Use the combination formula \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \) to calculate the number of ways to choose gaps.

### Step 4: Arrange the Girls
After choosing the gaps, we can arrange the n girls in those selected gaps. The number of ways to arrange n girls is given by \( n! \).

**Hint:** Remember that the arrangement of n distinct objects is \( n! \).

### Step 5: Combine the Results
Now, we can combine the results from the previous steps. The total number of arrangements is given by:
\[ \text{Total arrangements} = (m!) \times \left( \binom{m+1}{n} \right) \times (n!) \]

### Final Formula
Putting it all together, the total number of ways to arrange m boys and n girls such that no two girls are together is:
\[ m! \times \binom{m+1}{n} \times n! \]

### Summary
Thus, the final answer is:
\[ \text{Total arrangements} = m! \times \frac{(m+1)!}{n!(m+1-n)!} \times n! \]