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0 votes
10.3k views
in Permutation and Combinations by (15.7k points)

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.

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2 Answers

+2 votes
by (15.2k 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 (85 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

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