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 at least 5 boys sit between any two girls, we can follow these steps:
### Step 1: Understand the arrangement constraints
- We have 100 chairs.
- No girl can sit in the first or last chair.
- Between any two girls, there must be at least 5 boys.
### Step 2: Determine the positions for the girls
Since the girls cannot sit at either end, we can only place them in chairs 2 to 99. This gives us a total of 98 chairs available for the girls.
### Step 3: Calculate the minimum number of boys required
Since there are 10 girls and at least 5 boys must sit between each pair of girls, we need to calculate the number of boys required:
- Between 10 girls, there are 9 gaps.
- Each gap must have at least 5 boys.
- Therefore, the minimum number of boys required = 9 gaps × 5 boys/gap = 45 boys.
### Step 4: Calculate the remaining boys
We have a total of 90 boys. After placing the minimum required boys (45 boys), the remaining boys are:
- Remaining boys = 90 - 45 = 45 boys.
### Step 5: Determine the total gaps for boys
After placing the 10 girls, we can visualize the arrangement as follows:
- Gaps before the first girl, between the girls, and after the last girl.
- There are 11 gaps in total:
1. Before the first girl
2. Between each pair of girls (9 gaps)
3. After the last girl
### Step 6: Distribute the remaining boys
We need to distribute the remaining 45 boys into these 11 gaps. Let \( x_1, x_2, \ldots, x_{11} \) represent the number of boys in each of these gaps. The equation we need to solve is:
\[ x_1 + x_2 + x_3 + \ldots + x_{11} = 45 \]
with the conditions:
- \( x_2, x_3, \ldots, x_{10} \geq 0 \) (these are the gaps between girls)
- \( x_1 \geq 0 \) (gap before the first girl)
- \( x_{11} \geq 0 \) (gap after the last girl)
### Step 7: Use the stars and bars method
The stars and bars theorem states that the number of ways to distribute \( n \) indistinguishable objects (boys) into \( k \) distinguishable boxes (gaps) is given by:
\[ \binom{n+k-1}{k-1} \]
In our case, we have \( n = 45 \) and \( k = 11 \):
\[ \text{Ways to distribute boys} = \binom{45 + 11 - 1}{11 - 1} = \binom{55}{10} \]
### Step 8: Arrange the girls and boys
After determining the distribution of boys, we need to arrange the girls and boys:
- The 10 girls can be arranged among themselves in \( 10! \) ways.
- The 90 boys can be arranged among themselves in \( 90! \) 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:
\[ \binom{55}{10} \times 10! \times 90! \]
