# Math Help - this question is for the permutations unit...

1. ## this question is for the permutations unit...

At a banquet, four couples are sitting along one side of a table with men and women alternating.
a) How many arrangements are possible if each couple sits together? Explain your reasoning in sentences.
b) How many arrangements are possible if no one is sitting beside his or her partner?

2. Originally Posted by nithujaa
At a banquet, four couples are sitting along one side of a table with men and women alternating.
a) How many arrangements are possible if each couple sits together? Explain your reasoning in sentences.
treat each couple as one object. how many ways can you permute these objects to arrange them in a straight line? after you get that, multiply that answer by two, this covers the arrangements where the members of each couple may switch places.

b) How many arrangements are possible if no one is sitting beside his or her partner?
this is the total number of ways we can permute the 8 people, minus the number of arrangements where couples sit together (that is, minus your answer for part (a))

3. Ahem.

Jehvon,

If you take the number of permutations of 8 people and subtract out the number of arrangements where each pair sits together, you do not get the number of arrangements in which no pair sits together. Rather, the result is the number of arrangements in which at least one pair does not sit together.

Awkward

4. Originally Posted by awkward
Ahem.

Jehvon,

If you take the number of permutations of 8 people and subtract out the number of arrangements where each pair sits together, you do not get the number of arrangements in which no pair sits together. Rather, the result is the number of arrangements in which at least one pair does not sit together.

Awkward
indeed. the problem would be a lot harder then. it would be total arrangements minus arrangements where 1 or 2 or 3 or 4 couples sit together. that will be a pain. do you see an easier way?

5. The general problem is quite difficult, I think, but fortunately we only have to deal with the case of 4 couples.

Let's say the husbands are A, B, C, and D, and their wives are a, b, c, and d, respectively. If we take the case where the husbands sit down in the order ABCD and the leftmost person is a male (A), then there are only 3 possible arrangements:

A c B d C a D b
A c B d C b D a
A d B a C b D c

Start by observing that the person to the right of A must be c or d and it's easy to work out the possibilities.

Allowing for the 4! possible permutations of the husbands and the 2 possibilities of starting with a male or female in the leftmost seat, there are

$4! \times 2 \times 3$

possibilities in all.