Permutations O level Photograph Seating Arrangement Problem 
Here is the problem: Mum and Dad and their 4 children are to take a family photograph. They are to sit in a row containing 6 chairs.
Q: How many different ways of seating are there if Mum and Dad are to have exactly one of the children sitting between them?
ANS: 192

What I've tried: $3! \cdot 4C1 \cdot 2!$ because I guess the 3 remaining children would be moving from seat to seat, so, 3! and 4C1 because one child will be picked from 4 and 2! because Mum and Dad would be changing places with one of the children between them.
The result I am getting is $48$ and it's wrong.
 A: *

*Let's start with the M c D group possible placements. (You can think of it as gluing the three together, so that M c D becomes one piece to place in four chairs - the other three chairs being occupied by the remaining children):


*

*M c D c c c

*c M c D c c

*c c M c D c

*c c c M c D
Four ways to place parents with one child between them; but also Mum and Dad can switch places. So we have $4\cdot 2! = 8$ ways to place Mum and Dad.


*

*In other words, we need also to count the following four placements (pasting together D c M), in addition to the first four placements we listed:


*

*D c M c c c

*c D c M c c

*c c D c M c

*c c c D c M



*

*With respect to which child sits between Mum and Dad in the eight possible seatings of the parents above, we have $\dbinom 41 = 4$ ways to select which child sits between Mum and Dad.

*And lastly, we have $3!= 6$ ways to seat the remaining three children.
Therefore:
That gives us $\;4\cdot 2 \cdot 4 \cdot 3! = 192\;$ ways to seat the family given the required conditions.
