# Find the number of arrangements of 6 boys and 5 girls in a row so that 1. no two girls sit together.

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Find the number of arrangements of 6 boys and 5 girls in a row so that

1. no two girls sit together.

2. boys and girls occupy alternate positions.

by (13.4k points)

1. Since no two girls sit together, we have first arrange the 6 boys among themselves. This can be done in 6! ways.

 1 B 2 B 3 B 4 B 5 B 6 B 7

Now no two girls sit together if we place the girls in between boys. There are 7 places and it should be occupied by 5 girls, can be done in 7P5 ways.

Therefore the total number of ways is 6! × 7P5

= 720 × 7 × 6 × 5 × 4 × 3

= 1814400.

2. Boys and girls occupy alternate position can be done as follows.

First place the boys whose number is large.

 B 1 B 2 B 3 B 4 B 5 B

Boys can be arranged in 6! ways. The place between boys can be filled by 5 girls, can be done in 5! ways.

Therefore the total number of ways is 6! × 5! = 720 × 120 = 86400.

+1 vote
by (70 points)
edited
1. Since no two girls sit together, we have to first arrange the 6 boys among themselves. This can be done in 6! ways.
_ B1 _ B2 _ B3 _ B4 _ B5 _ B6 _ B7 _

Now, no two girls sit together if we place the girls in the places kept empty in above illustration with ‘_’s. There are 7 such places and it should be occupied by 5 girls, which can be done in 7Pways.

Hence, the required total number of ways
= 6! × 7P5
= 720 × 7 × 6 × 5 × 4 × 3
= 1814400

2.
Boys and girls can occupy alternate position as follows.

B1B2 _ B3 _ B4 _ B5 _ B6 _ B7

First we place a boy as they are more in number. Thus, leaving a space each for a girl, we may arrange 6 boys, keeping a total of 5 empty spaces between them.

Boys can be arranged in 6! ways. The place between boys can be filled by 5 girls in 5! ways.

Hence the required total number of ways
= 6! × 5!
= 720 × 120
= 86400