In how many ways can 4 men and 3 women can arrange with a condition that each men should not sit together and they must be in the order of their age.
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one woman will set between two men and one men with other women and one men will set one woman
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@ranshsangwan
Given that the total number of people = 7
First we select 4 positions for the men to occupy.
These positions could be chosen in 7C4 ways = 7!/4!(7-4)! = (1 x 2 x ... x 7)/(1 x 2 x 3 x 4)(1 x 2 x 3)
= (5 x 6 x 7)/(1 x 2 x 3) = 35 ways.
This leaves 3 positions for the women, and the women can be rearranged amongst themselves in 3! ways = 6 ways.
The men, of course, can be in only one order amongst themselves, so the overall number of ways they could line up is given by 35 x 6 = 210 ways.
Hence, the answer is 210
Given that the total number of people = 7
First we select 4 positions for the men to occupy.
These positions could be chosen in 7C4 ways = 7!/4!(7-4)! = (1 x 2 x ... x 7)/(1 x 2 x 3 x 4)(1 x 2 x 3)
= (5 x 6 x 7)/(1 x 2 x 3) = 35 ways.
This leaves 3 positions for the women, and the women can be rearranged amongst themselves in 3! ways = 6 ways.
The men, of course, can be in only one order amongst themselves, so the overall number of ways they could line up is given by 35 x 6 = 210 ways.
Hence, the answer is 210
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