perms and combs 2 (2 Viewers)

[jnney]

a rectangular table has 6 seats, 4 being on one side and 2 on the opposite side. in how many ways can 6 people be seated if A and B must be on opposite sides?

 

[deswa1]

Note: I haven't done this topic, so I'm going to do it logically (might be wrong):

Either A can be on the side with 2 or 4 seats. If A is on the side with 2, then B has four seats to choose from. Everyone else can sit anywhere giving 4x4!.
If A is on the side with 4, then B has 2 seats to choose from and everyone else can sit anywhere giving 2x4!.
Adding together gives 6x4! or 144.

Can someone confirm if that's right

 

[jnney]

Answer is 384!

 

[deswa1]

All right, I know what I did wrong. When A sits on the side that has 2 seats, he can sit in 2 different positions, giving 2 for him, 4 for B and 4! for everyone else (8x4!). Similarly, when A sits on the side with 4, he can sit in four seats, B in 2 and the rest in 4!. Thus the total is 8x4!x2 or 384.

 

[Shadowdude]

Let the side with 4 seats be Side X, and let the side with 2 seats be Side Y.

Case 1: A on Side X and B on Side Y

1. Place A on a seat on Side X ... 4 ways
2. Place B on a seat on Side Y ... 2 ways
3. Place the four other people on the four other seats ... 4! ways
Total: 8*(4!)

Case 2: A on Side Y, B on Side X

1. Place A on a seat on Side Y ... 2 ways
2. Place B on a seat on Side X ... 4 ways
3. Place the four other people on four other seats ... 4! ways
Total: 8*(4!)

Answer: 2(8*(4!)) = 384

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Simple tip on how to do this, you set it out as I have done. It's logical and you'll be able to spot mistakes easier.