Arrangements and Derangements

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**MATHDUDE2** Hi,

I'm having trouble solving this problem. I'm not sure how to go about part b) and onward. Could someone please help me by explaining what must be done?

Question: At a party, four couples, are sitting along one side a table with men and women alternating.

a) How many seating arrangements are possible for these eight people?

b) How many arrangements are possible if each couple sits together? Explain your reasoning.

c) How many arrangements are possible if no one is sitting beside his or her partner?

d) Explain why the answers from parts b) and c) do not add up to the answer from part a).

Thank you

Part (b)

Suppose the men are seated first. This can be done in $\displaystyle 2n$ ways (using Matt Westwood's definition of $\displaystyle n$). Then there is only one seat in which each of the women can sit, if each is to sit by her partner. So there's the answer: $\displaystyle 2n$.

Part (c)

Suppose, as in part (b), we seat the men first (in $\displaystyle 2n$ ways as before). Then* none *of the women can occupy the seat that she did in part (b). Not only that, but she cannot occupy a seat (if there is one) on the *other *side of her partner. In fact, once the men are seated, there are then only 3 ways in which the women can be seated if no couple is seated together. Here are the three arrangments:

Suppose the men are A, B, C and D, and their partners a, b, c, and d respectively. Then suppose that the men are seated like this:

A * B * C * D *

Then 'a' can occupy any one of three positions as follows:

- Between B and C: A * B a C * D *

In this case, d must sit between A and B: A d B a C * D *

and therefore c here: A d B a C * D c.

So the only possible arrangement is A d B a C b D c

- Between C and D: A * B * C a D *

Following a similar argument, the only possible arrangement now is A c B d C a D b

- Beyond D: A * B * C * D a

and the only arrangment now is A c B d C b D a

Clearly the same argument will apply whatever the position occupied by the four men. Thus the number of arrangements in which no couple is together is $\displaystyle 3 \times 2n = 6n$.

Part (d)

The only way in which the answers to (b) and (c) would add together to make (a) is if (b) and (c) were the only possibilities. But that's not so, is it? What else might happen?

Grandad