Find the ways in which no husband can sit next to another husband A group of 6 people consisting 3 married couples are to be seated together in a straight row. How many different ways are there of seating the 6 people if no husband is to sit next to another husband?
Since the husbands can't sit next to each other, I decided to place the wife in between the husband and arrange them.
H W H W H W 
3! * 3! = 36. 
Apparently, the answer is 144. Can anyone explain what went wrong?
 A: The positions that can ocuppy the husbands are: $135, 136, 146, 246$
A: With some work, show that there are four possible arrangements:
WHWHWH  (and it's reflection)
HWHWWH  (and it's reflection).
Can you finish the problem from here?
A: 
Can anyone explain what went wrong?

The positions of the husbands and wives do not need to alternate since the wives may sit in adjacent seats.  For instance, the seating arrangement $H_1W_1H_2W_2W_3H_3$ is permissible since no two of the husbands sit in adjacent seats.

A group of $6$ people consisting of $3$ married couples are to be seated together in a straight row. How many different ways are there of seating the $6$ people if no husband is to sit next to another husband?

Method 1:  We arrange the wives, then insert the husbands among the wives.
Arrange the husbands in some order, say alphabetically, then hand each of them a chair.
Lay out three chairs for the wives.  The wives can arrange themselves in those chairs in $3!$ ways.  For each such arrangement arrangement, there are four spaces in which we can place the husbands, two between successive wives and two at the ends of the row.
$$\square W_1 \square W_2 \square W_3 \square$$
To ensure that the husbands do not sit in adjacent positions, they must choose three of these four spaces.  The first husband has four choices where to place his chair, the second has three, and the third has two.  Hence, the number of permissible seating arrangements is 
$$3! \cdot 4 \cdot 3 \cdot 2 = 3!4!$$ 
Method 2:  We use the Inclusion-Exclusion Principle.
The six people can be arranged in $6!$ ways.  From these, we must exclude those arrangements in which two husbands sit in adjacent seats.
A pair of husbands sit in adjacent seats:  There are $\binom{3}{2}$ ways to choose a pair of husbands who sit in adjacent seats.  This gives us five objects to arrange, the block consisting of the pair of husbands and the other four people.  The objects can be arranged in $5!$ orders.  Within the block, the two husbands can be arranged in $2!$ orders.  Hence, there are 
$$\binom{3}{2}5!2!$$
arrangements in which two husbands sit in adjacent seats.
However, 
$$6! - \binom{3}{2}5!2! = 0$$
Clearly, we have subtracted too much since the seating arrangement stated above shows it is possible to seat the three couples so that no two husbands sit in adjacent seats.
The reason we have subtracted too much is that we have subtracted those cases in which all three husbands sit together twice, once when we counted the leftmost and middle husband as our designated pair and once when we counted the rightmost and middle husband as our designated pair.  We only want to subtract them once, so we need to add those cases to the total.
Two pairs of adjacent husbands:  This can only occur if all three husbands sit in a block.  This gives us four objects to arrange, the block of three husbands and the three wives.  The four objects can be arranged in $4!$ ways.  Within the block, the husbands can be arranged in $3!$ ways.  Hence, there are 
$$4!3!$$
seating arrangements with two pairs of adjacent husbands.
Total:  By the Inclusion-Exclusion Principle, the number of permissible seating arrangements is 
$$6! - \binom{3}{2}5!2! + 4!3!$$
