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 Apr 16, 2024

#1**0 **

We can solve this problem by considering different cases for how the siblings are arranged across the two rows.

Case 1: Splitting Each Pair

In this case, we have one child from each pair in each row. There are 3!=6 ways to arrange the children in each row (since sibling order within a row doesn't matter). However, we've overcounted since we haven't considered which sibling from each pair sits in which row. So, for each arrangement, we need to multiply by 23 to account for the two choices (sibling 1 or sibling 2) for each of the three pairs. This gives us 6⋅23=48 arrangements for this case.

Case 2: Keeping Pairs Together

Here, both siblings from one or two pairs sit in the same row. There are two sub-cases to consider:

Two Pairs Together: There are (23)=3 ways to choose which two pairs will sit together (the third pair will be split). Once chosen, there are 2!=2 ways to arrange the two siblings within each pair that sits together, and again 2 ways to decide which row the combined pair sits in. Finally, there are 2!=2 ways to arrange the remaining split pair in the other row. This gives a total of 3⋅2⋅2⋅2=24 arrangements for this sub-case.

One Pair Together: There's only one way to choose which pair will sit together. Within that pair, there are 2! ways to arrange them. There are then 2 ways to decide which row the pair sits in, and 3!=6 ways to arrange the remaining two pairs (who must be split) in the other row. This gives a total of 1⋅2⋅2⋅6=24 arrangements for this sub-case.

Adding the sub-cases of Case 2, we get a total of 24+24=48 arrangements.

Total Arrangements

Summing the arrangements from both cases, we get a total of 48(Case1)+48(Case2)=96 ways to seat the siblings.

Boseo Apr 16, 2024