To solve this problem, we can use the concept of permutations and combinations.

Step 1: First, arrange the 6 boys in a row. This can be done in 6! (factorial) ways.

Step 2: Now, there are 7 places (6 gaps between the boys and 1 at each end) where the 4 girls can be seated such that no two girls are together. This can be done in 7P4 ways (permutations of 7 items taken 4 at a time).

Step 3: Multiply the number of ways from step 1 and step 2 to get the total number of ways the boys and girls can be seated.

So, the total number of ways = 6! * 7P4.

Question

To solve this problem, we can use the concept of permutations and combinations.

Step 1: First, arrange the 6 boys in a row. This can be done in 6! (factorial) ways.

Step 2: Now, there are 7 places (6 gaps between the boys and 1 at each end) where the 4 girls can be seated such that no two girls are together. This can be done in 7P4 ways (permutations of 7 items taken 4 at a time).

Step 3: Multiply the number of ways from step 1 and step 2 to get the total number of ways the boys and girls can be seated.

So, the total number of ways = 6! * 7P4.

Knowee AI · Accepted Answer

To solve this problem, we can use the concept of permutations and combinations.

Step 1: First, arrange the 6 boys in a row. This can be done in 6! (factorial) ways.

Step 2: Now, there are 7 places (6 gaps between the boys and 1 at each end) where the 4 girls can be seated such that no two girls are together. This can be done in 7P4 ways (permutations of 7 items taken 4 at a time).

Step 3: Multiply the number of ways from step 1 and step 2 to get the total number of ways the boys and girls can be seated.

So, the total number of ways = 6! * 7P4.

In how many ways 4 girls and 6 boys can be seated in a row so that no two girls are together?

Question

In how many ways can 4 girls and 6 boys be seated in a row so that no two girls are together?

Solution

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