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7 men and 10 women must be seated in a row, with the women taking the even-numbered seats. How many agreements of this kind are possible?

Option: 1

10! 4!


Option: 2

9! 5!


Option: 3

9! 7!

 


Option: 4

10! 7!


Answers (1)

best_answer

Given that,

There are 7 men and 10 women.

The total number of people is 17.

Women only occupy even positions. Women are limited to 10 even spaces.

Thus, 10 women can sit in 10 positions.

^{10}P_{10}=\frac{10!}{(10-10!)}

^{10}P_{10}=10!

7 men can sit in 7 positions.

Thus,

^{7}P_{7}=\frac{7!}{(7-7!)}

^{7}P_{7}=7!

Therefore, the possible number of arrangements is 10! 7!.

 

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vinayak

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