# Thread: in how many ways can...

1. ## in how many ways can...

In how many ways can you seat 4 couples at a round table if:
1) each couple sits together?
2) no couples sits together?

I got 12 for the first...I highly doubt it's right though. Any tups? I don't want you to actually solve it all out for me, but pointing me in the right direction would be nice
thanks!
I know that: # of circular combinations = # of linear combinations / # of circle combinations that can occur from each linear combination.
I just can't really figure out exactly how to apply that to these problems...
=/

2. In how many ways can you seat 4 couples at a round table if:
1) each couple sits together?
2) no couples sits together?

you have 8 people. 8 chairs

1. If the couple has to sit together then...the first chair has 8 possible ways, but the next one can only have 1 possible way which is the persons significant other. You have 6 chairs left and the 3rd chair now has 6 possible ways, and the next one to it has only 1...and etc... so...

8 * 1 * 6 * 1 * 4 * 1 * 2 * 1 = 384 ways.

2. the total possible ways to seat these 4 couples with no restrictions...
8! = 40320

if no couples sits together....then you take the total possible ways minus the possible ways of them sitting together.

8! - 384 = 39936

3. Hello, Sconts!

Here's part (1) . . .

In how many ways can you seat 4 couples at a round table if:
(1) each couple sits together?
(2) no couples sits together?

1) If the couples must be adjacent, duct-tape the pairs together.
Then we have four "people" to arrange around a round table.

The first couple can sit anywhere
There are 3! = 6 ways to seat the other three couples.

But each couple can be seated in two ways: Aa or aA.
Hence, there are: 2^4 = 16 possible orderings.

Therefore, there are: .6 × 16 .= .96 possible seatings.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Using smoothi's approach . . .

The first person can sit anywhere; call that person/position #1.
. . Let us move clockwise around the table.
Position #2 must be occupied by 1's spouse: 1 choice.

Position #3 can be occupied by any of the six remaining people: 6 choices.
Position #4 must be occupied by 3's spouse: 1 choice.

Position #5 can be occupied by any of the four remaining people: 4 choices.
Position #6 must be occupied by 5's spouse: 1 choice.

Position #7 can be occupied by either of the two remaining people: 2 choices.
Position #8 must be occupied by 7's spouse: 1 choice.

Hence, there are: .1 x 6 x 1 x 4 x 1 x 2 x 1 .= .48 ways.

If we move counterclockwise around the table, we find 48 more arrangements.
. . These are mirror-images of the first 48 arrangements.

Therefore, there are: .48 + 48 .= .96 possible seatings.