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8 boys and 4 girls will be seated in two separate rows of 6 chairs each so that two specific girls are always together and no girls are in the same row. How many different positions can they take? 

Option: 1

16\times 10!

 

 


Option: 2

18\times 10!

  

  

 


Option: 3

20\times 10!  

 


Option: 4

22\times 10! 

 


Answers (1)

best_answer

Two girls will always be together.

So, first, we choose one row for these two girls to sit in, ways = 2.

We can now choose 2 adjacent chairs in four different ways, and the girls can sit on them in 2 different ways.

Seating for these two girls is2\times 4\times 2=16.

All of the girls should not be in the same row, so at least one girl should be in a row separate from the remaining two girls.

So, from the remaining two girls, we choose one to sit in the second row, ways = 2.

Now, in the second row, we choose one of 6 chairs to seat the chosen girl, ways = 6.

Seating for the third girl is  5\times 2=10 

The remaining nine can now be arranged in 9!

Therefore, the total number of ways is 16\times 10\times 9!=16\times 10!.

Posted by

HARSH KANKARIA

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