Question from http://www.sgforums.com/forums/2297/topics/302719
Four men, three women and two children are to be seated in a row under the following restrictions : each child must be seated between 2 men and no women must sit next to each other. Find the number of possible seating arrangements.
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Answer:
We have 3 cases
W (M C M) W (M C M) W
W (M C M C M) W (M) W
W (M) W (M C M C M) W
for the first case ---- number of ways = 3! × 4C2 × 2! × 2 × 2! = 288
[permute the 3 women first (3!), then choose 2 men out of 4 (4C2), and permute the men in 1 set (2!), then choose 1 child out of 2 (2), and for the other set u can permute the men also (2!)]
for the second and third case ---- number of ways = 3! × 4C3 × 3! × 2! × 2 = 576
[permute the 3 women first (3!), then choose 3 men out of 4 (4C3), and permute the men in 1 set (3!), then permute the 2 children (2!), and swap the 2 bracketed sets (2) to permutate between the 2nd and 3rd case
Adding them together, total number of ways = 576 + 288 = 864
