In how many ways can three pairs of siblings from different families be seated in two rows of three chairs, if siblings may sit next to each other in the same row, but no child may sit directly in front of their sibling?

ABJeIIy Jun 3, 2024

#1**+2 **

One way to deal with the restriction is to solve the problem as if it didn't exist and then subtract the solutions we can't have.

There are 6 children and 6 seats. The 1st child has 6 choices. For each of those 6 options, the 2nd child has 5 choices, because the 1st child occupies a chair. For each of those 5 options, the 3rd child has 4 choices, and so on. So, there are 6*5*4*3*2*1 options. One way to see this is to draw a counting tree. 6*5*4*3*2*1 is more commonly known has 6 factorial, or 6!. 6! = 720, which is the number of ways you can sit the children without restrictions. However, no two siblings can sit in front of their sibling. This removes 3 options, so the answer is 720 - 3 = \(\boxed{717}\).

PurpleWasp Jun 3, 2024