5 Men And 4 Women Are To Be Seated In A Row So That The Women Occupy

1.	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?
Answer» 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. 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.

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?

Answer»

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.

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.