Question 1131553: In how many ways can 6 people be seated in a row if
(a) Persons A and B must sit next to each other?
(b) There are 3 men and 3 women, no 2 men and 2 women can sit next to each other?
Answer by greenestamps(13203)   (Show Source):
You can put this solution on YOUR website!
(a) A and B must sit together.
(1) Consider A and B as a "single person". Then there are 5 "people" to be seated; this can be done in 5! = 120 ways.
(2) In the group of A and B, the two can be seated in either of 2 different orders.
ANSWER: 120*2 = 240 ways.
(b) 3 men and 3 women; no 2 men and no 2 women can sit next to each other (i.e., the seating must alternate men and women).
Consider seating the people one at a time, from the beginning of the row.
(1) Any of the 6 people can be seated first: 6 choices
(2) Without loss of generality, suppose the first person seated was a woman. Then the second person seated must be one of the 3 men: 3 choices
(3) The third person seated must be one of the remaining 2 women: 2 choices
(4) The fourth person must be another man: 2 choices
(5) The fifth person must be the 3rd woman: 1 choice
(6) The sixth person must be the 3rd man: 1 choice
The total number of ways to seat the 6 people is the product of the numbers of choices for each seat.
ANSWER: 6*3*2*2*1*1 = 72 ways
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