# i need some idea in this problem of permutations

In how many different ways 8 people can be seated in a row if 1. there are no restriction on the seating. 2. persons A and B must next to each other 3. there are 4 men and 4 women and no two men or 2 women can sit next to each other 4. there are 5 men and they must sit next to each other 5. there 4 married couple and each couples and each couple must sit togther. I have tried and solve some of them like 1, 2 but not clear with other

• What do your attempts look like? Can you walk us through your thought process? – John Engbers Jun 22 '13 at 0:55

Hints: For #2, there are $7$ possible pairs of seats in which we can place persons $A$ and $B$ (why?). How many ways can we arrange persons $A$ and $B$? How many ways can we arrange everyone else?

For #3, there are only two possible gender arrangements (why?). Once we've chosen our arrangement, how many ways can the men be ordered? What about the women?

For #4, there are $4$ places we can put our "man cluster" (why?). Once we've decided where the "man cluster" will go, how many ways can we arrange the women? How many ways can we arrange the men in the cluster?

For #5, there are $4$ possible "married couple spots." How many ways can we assign these spots? How many ways can we arrange each married couple in their assigned spot?

(1) You know that the answer is $8!$.

(2) Since A and B want to be close let's put them in a bag. Now we have $7$ "people" ($6$ real ones and the bag). These can be arranged in $7!$ ways. Let A and B out of the bag. They can sit in either order, so the total is $(2)(7!)$.

Or else, but this is less fun, there are $7$ possible pairs of seats that can be used by the two fussy people. For each such pair, A and B can be seated in $2$ ways. And now that they are seated, the remaining $6$ empty spaces can be filled in $6!$ ways, for a total of $(7)(2)(6!)$.

(3) The only allowed seatings are WMWMWMWM and MWMWMWMW, For the first type, the women can be arranged in their $4$ allowed spaces in $4!$ ways, and for each such choice the men can be arranged in $4!$ ways. That gives $(4!)(4!)$. Double to take care of the second type of arrangement.

(4) Put all the men in a bag. Then we have $4$ "people," who can be arranged in $4!$ ways. Let the men out of the bag. They can arrange themselves in $5!$ ways for a total of $(4!)(5!)$.

(5) We need more bags, $4$ of them. Put each couple in a bag. The bags can be arranged in $4!$ ways. Let the people out of the bags. The contents of each bag can arrange themselves in either order, for a total of $(4!)(2^4)$.