There are four couples to be seated in a row with $8$ seats.

Question

the number of ways they can seat

(a) without any restriction is $8!$.

(b) each couple is seated together is $4!$.

(c) the males and females have to seat together is $(4! \times 4!) \times 2$.

Is the solution correct?

Saketh Malyala · Accepted Answer · 2018-10-07 05:52:43Z

1

Your answer for b is a bit off.

If you treat each couple as a block, then you can arrange the blocks in $4!$ ways.

However, you can arrange each couple INSIDE the block in $2$ ways.

There are $4! \cdot 2^4$ ways to have each couple sitting together.

answered Oct 7, 2018 at 5:52

Saketh Malyala

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score 0 · Accepted Answer · 2018-10-30 07:26:02Z

0

your part $b$ is not right

couples can also permute between themselves in $2!$ ways so better multiply your answer in part$\ b\ $by $(2!)^4$ rest is fine

answered Oct 7, 2018 at 5:56

user590550

$\begingroup$ Your answer is also incorrect. As you said, each couple can be permuted in $2!$ ways. Since there are four couples, we need to multiply by $(2!)^4$. $\endgroup$
– N. F. Taussig
Oct 7, 2018 at 13:08

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