In how many ways can 4 boys and 5 girls can stand such that no two boys are together.

Answer

To solve the problem of arranging 4 boys and 5 girls such that no two boys are together, we can follow these steps:

### Step 1: Arrange the Girls
First, we will arrange the 5 girls. The number of ways to arrange 5 girls is given by the factorial of the number of girls:

\[ \text{Ways to arrange girls} = 5! = 120 \]

**Hint:** Remember that the factorial of a number \( n \) (denoted as \( n! \)) is the product of all positive integers up to \( n \).

### Step 2: Identify Positions for Boys
Once the girls are arranged, we can visualize the arrangement. The girls create gaps where the boys can be placed. For 5 girls, the arrangement looks like this:

\[ G \_ G \_ G \_ G \_ G \]

Here, \( G \) represents a girl and \( \_ \) represents a potential position for a boy.

In total, there are 6 gaps (before the first girl, between the girls, and after the last girl) where the boys can be placed:

\[ \_ G \_ G \_ G \_ G \_ G \_ \]

**Hint:** Count the gaps created by arranging the girls to determine where the boys can be placed.

### Step 3: Choose Positions for Boys
We need to choose 4 out of these 6 gaps to place the boys. The number of ways to choose 4 gaps from 6 is given by the combination formula:

\[ \text{Ways to choose gaps} = \binom{6}{4} = 15 \]

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

### Step 4: Arrange the Boys
After selecting the gaps, we need to arrange the 4 boys in the chosen gaps. The number of ways to arrange 4 boys is given by:

\[ \text{Ways to arrange boys} = 4! = 24 \]

**Hint:** Similar to arranging girls, use the factorial to find the number of arrangements for the boys.

### Step 5: Calculate the Total Arrangements
Now, we can calculate the total number of arrangements by multiplying the number of ways to arrange the girls, the number of ways to choose the gaps for the boys, and the number of ways to arrange the boys:

\[ \text{Total arrangements} = (5!) \times \left(\binom{6}{4}\right) \times (4!) \]
\[ = 120 \times 15 \times 24 \]

Calculating this gives:

\[ = 120 \times 15 = 1800 \]
\[ = 1800 \times 24 = 43200 \]

### Final Answer
Thus, the total number of ways in which 4 boys and 5 girls can stand such that no two boys are together is:

\[ \boxed{43200} \]