Vtyjkyuk

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.

• 
  

  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):

  
  
  
  1. 
     M c D c c c
  
  
  2. c M c D c c
  
  
  3. c c M c D c
  
  
  4. 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 $4cdot 2! = 8$ 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:

  
  
  
  1. 
     D c M c c c
  
  
  2. c D c M c c
  
  
  3. c c D c M c
  
  
  4. 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 $;4cdot 2 cdot 4 cdot 3! = 192;$ ways to seat the family given the required conditions.