problem with Combination and Permutation Four married couples have bought 8 seats in a row for a concert.
In how many ways can they be seated:
a)if each couple is to sit together?
(8)(1)(6)(1)(4)(1)(2)(1)
b)if all men sit together?
n-k+1=8-4+1=5
so, (5)(4! 4!)=5! 4!
c)If no man sits next to his wife ?(This is a non-trivial questions)
8*6*6*4*4*2*2*1=18432 ways
could you please check it for me ?

a) and b) are correct.
For c), we will use Inclusion/Exclusion. There are $8!$ arrangements without restriction. From this we need to subtract the number of bad arrangements, where at least one couple are next to each other.
First we count the number of arrangements where Couple A are together. Tie them together with rope. There are then $7$  objects to be arranged. This can be done in $7!$ ways. Now untie Couple A. They can occupy $2$ different positions, for a total of $2\cdot 7!$. Multiply by $\binom{4}{1}$ for the number of ways to choose a couple. Our first estimate of the number of bad arrangements is $\binom{4}{1}\cdot 2\cdot 7!$.
However, this grossly overcounts the number of bad arrangements, for it double-counts, among others, the arrangements where Couple A and B are both next to each other. An analysis similar to the previous one shows that there are $2^2\cdot 6!$ arrangements in which Couple A and B are together. Thus our adjusted count for the number of bad arrangements is $\binom{4}{1}\cdot 2\cdot 7!-\binom{4}{2}\cdot 2^2\cdot 6!$.
However, we have subtracted too much, for we have subtracted one too many times, for example, the arrangements where couples A, B, C are together. Adjusting for this gives the adjusted count $\binom{4}{1}\cdot 2\cdot 7!-\binom{4}{2}\cdot 2^2\cdot 6!+\binom{4}{3}\cdot 2^3\cdot 5!$. 
A final adjustment needs to be made, we must subtract the $\binom{4}{4}\cdot 2^4\cdot 4!$ arrangements in which all couples are together. This will give us the total number of bad arrangements, and now we are nearly finished.
