In a class of 4 boys and 3 girls, they are required to sit in a row such a way that no two girls can sit together. Compute in how many different ways they can sit together

To solve this problem, we need to arrange 4 boys and 3 girls such that no two girls sit together. Here's a step-by-step breakdown:

### Step 1: Arrange the boys
First, arrange the 4 boys. The number of ways to arrange 4 boys is the number of permutations of 4 boys, which is:

\[
4! = 4 \times 3 \times 2 \times 1 = 24
\]

### Step 2: Determine possible seats for the girls
Once the boys are seated, the girls must sit in the available spaces between the boys. Consider the boys as forming a row with spaces around them:

\[
\_ \, B \, \_ \, B \, \_ \, B \, \_ \, B \, \_
\]

There are 5 available positions (denoted by underscores) where the girls can sit. Since no two girls can sit together, the girls must occupy 3 out of these 5 available spaces.

### Step 3: Select the seats for the girls
The number of ways to choose 3 spaces out of the 5 available positions is given by the combination formula:

\[
\binom{5}{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10
\]

### Step 4: Arrange the girls in the selected seats
After selecting 3 positions for the girls, we can arrange the 3 girls in those 3 positions in \(3!\) ways:

\[
3! = 3 \times 2 \times 1 = 6
\]

### Step 5: Total number of arrangements
Now, we multiply the number of ways to arrange the boys, select the positions for the girls, and arrange the girls in those positions. Therefore, the total number of arrangements is:

\[
4! \times \binom{5}{3} \times 3! = 24 \times 10 \times 6 = 1440
\]

Thus, the total number of ways to arrange the 4 boys and 3 girls such that no two girls sit together is:

\[
\boxed{1440}
\]

Would you like further details or have any questions?

Here are some related questions you might find interesting:
1. What if there were 5 boys and 4 girls instead, still ensuring no two girls sit together?
2. How would the arrangement change if we allowed exactly two girls to sit together?
3. What is the probability that a random arrangement of these 7 students (no seating restriction) has no two girls sitting together?
4. How would the arrangement change if boys also can't sit together?
5. How many ways can they sit if there is no restriction on seating?

**Tip**: When dealing with seating arrangement problems involving restrictions, always separate the tasks (e.g., arranging people, selecting seats) and then combine them using multiplication.

In a class of 4 boys and 3 girls, they are required to sit in a row in such a way that no two girls can sit together. Compute in how many different ways they can sit together.

Learn how to compute the number of ways to seat 4 boys and 3 girls in a row such that no two girls sit together. This problem involves key mathematical concepts such as permutations, combinations, and the factorial function, and is suitable for high school students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation