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5 men and 4 women are to be seated in a row so that the women occupy the even places . How many such arrangements are possible?

Correct Answer: 2880

Explanation:

There are total 9 places out of which 4 are even and rest 5 places are odd.


 


4 women can be arranged at 4 even places in 4! ways.


 


and 5 men can be placed in remaining 5 places in 5! ways.


 


Hence, the required number of permutations  = 4! x 5! = 24 x 120 = 2880


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