In how many ways can 14 people be seated in a row if:
a.) there are 7 men and 7 women and no two men or two women sit next to each other?
My attempt: Since no two men or women can sit next to each other I calculated $(7-1)! \cdot (7-1)! = 518400$
b.) there are 8 men and they must sit next to one another?
My attempt: If 8 men must sit next to one another, then there are 6 women left. What I did was calculate $(6 + 1)! = 5040$
Is this the correct approach?
Best Answer
Recall that the number of permutations of $n$ distinct objects is $n!$ (not $(n-1)!$ or $(n+1)!$).
For (a). The 7 men and the 7 women are alternated and the row can start with a man or with a woman (2 cases). Therefore the number of ways is $$2\cdot 7!\cdot 7!$$
For (b). We have a block of 8 men and 6 women, and the last man in the block can be in position $8$th to $14$th (7 cases). Therefore the number of ways is $$7\cdot 8!\cdot 6!$$