+355 In how many ways can three pairs of siblings from different families be seated in two rows of three chairs, if siblings may not sit next to each other in the same row?
+535 There are two cases to consider:
Case 1: Siblings of one pair sit in the same row
There are 3 ways to choose which pair of siblings will have siblings sitting in the same row (either the first row, the second row, or both).
Once a pair is chosen, there are 2 ways to arrange the siblings within that pair (who sits in the first seat and who sits in the second).
For the remaining two pairs, there are only 4 choices for the first seat in the row they occupy (since two seats are already taken by siblings from the chosen pair). Then there are 3 choices for the second seat (since one child has already been seated), for a total of 4 * 3 = 12 ways to arrange the children in that row.
Therefore, the total number of arrangements for Case 1 is 3 (ways to choose the pair) * 2 (arrangements within the pair) * 12 (arrangements in the other row) = 72.
Case 2: Siblings of each pair sit in different rows
There are 6 choices for the child in the first seat of the first row. It doesn't matter which sibling takes it, so suppose child A sits there (denoted by A).
Then there are 4 choices for the second seat (child B, child C, child D, or child E). The last seat in the first row cannot be filled by a sibling of child A, so it must be filled by one of the remaining two children (child D or child E).
This leaves a pair of siblings (let's say child B and child C) and one person (child D or child E) for the second row. So the only order that will work is child B - empty seat - child C. There are 2 choices for which sibling (B or C) sits in the first seat of the second row.
Therefore, the total number of arrangements for Case 2 is 6 (choices for child in first seat of first row) * 4 (choices for second seat) * 2 (choices for who sits first in the second row) = 48.
Adding the ways from both cases, there are 72 (Case 1) + 48 (Case 2) = 120 ways to seat the siblings.
