I found this solution somewhere. It was helpful for me.

Solution:

let the couple be aa,bb and the person be x

first case fix X at position no. 1. Then the adjacent seat will have 4 choices i.e 4 ways, let it be a xa_ _ _

Now the next seat has 2 choices(bb).... -> xab_ _

and now next two seat will have 1 choice each so that choice in this case is 1*4*2*1*1=8.

Same no of choices will be true when x takes the last seat i.e ababx

Till here 8+8=16

case three when x is in second position...

_x_ _ _

For first position 4 choices ax_ _ _

Now the third postion (i.e is next to x) will have two choice from bb in order to satisfy the condition axbab

So 4*2*1*1*1=8.

Similarly for abaxb eight ways

total=16+8+8=32

case five When x is in centre

_ _ x _ _

1st seat has 4 choices ; second seat has 2 choices ;4th seat has 2 choices ; last 1 choice

4*2*1*2*1=16

tot=48

probability=48/120=2/5
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Please +1 KUDO if my post helps. Thank you.