Find the number of ways in which 10 girls and 90 boys can sit in a row having 100 chairs such that no girls sit at then either end of the rwo and between any two girls, at least five boys sit.

Answer

To solve the problem of arranging 10 girls and 90 boys in a row of 100 chairs such that no girls sit at either end and there are at least 5 boys between any two girls, we can follow these steps:

### Step 1: Understand the arrangement constraints
- We have 10 girls (G) and 90 boys (B).
- No girl can sit at either end of the row, meaning the first and last chairs must be occupied by boys.
- Between any two girls, there must be at least 5 boys.

### Step 2: Calculate the minimum number of boys required
Since there are 10 girls, there will be 9 gaps between them. To satisfy the condition of having at least 5 boys between each pair of girls, we need:
- 9 gaps × 5 boys = 45 boys.

### Step 3: Determine the remaining boys
After placing 45 boys between the girls, we have:
- Total boys = 90
- Boys used = 45
- Remaining boys = 90 - 45 = 45 boys.

### Step 4: Visualize the arrangement
Now we can visualize the arrangement:
- G B B B B B G B B B B B G B B B B B G B B B B B G B B B B B G B B B B B G B B B B B G B B B B B G
- This arrangement has 10 girls and 45 boys already placed, leaving us with 45 boys to be arranged freely.

### Step 5: Identify the gaps for remaining boys
After placing the 10 girls and the 45 boys, we can identify the gaps where the remaining boys can be placed:
- There are 11 gaps (1 before the first girl, 9 between the girls, and 1 after the last girl).

### Step 6: Distributing the remaining boys
We need to distribute the remaining 45 boys into these 11 gaps. This can be modeled as a problem of distributing indistinguishable objects (boys) into distinguishable boxes (gaps).

### Step 7: Use the stars and bars theorem
The number of ways to distribute \( n \) indistinguishable objects into \( r \) distinguishable boxes is given by the formula:
\[ \text{Number of ways} = \binom{n + r - 1}{r - 1} \]
In our case:
- \( n = 45 \) (remaining boys)
- \( r = 11 \) (gaps)

Thus, the number of ways to distribute the boys is:
\[ \binom{45 + 11 - 1}{11 - 1} = \binom{55}{10} \]

### Step 8: Arranging the girls
The 10 girls can be arranged among themselves in \( 10! \) ways.

### Step 9: Calculate the total arrangements
The total number of arrangements is given by:
\[ \text{Total arrangements} = \binom{55}{10} \times 10! \times 90! \]

### Final Answer
Thus, the total number of ways in which 10 girls and 90 boys can sit in a row under the given conditions is:
\[ \binom{55}{10} \times 10! \times 90! \]