+136 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?
+1145 We can solve this problem by considering two cases:
Case 1: One family has siblings in each row
There are 3 families (let's call them A, B, and C). In this case, we need to place one child from each family in each row. There are 3 choices for the first seat in the first row (child from any family).
Once the first seat is filled, there are 2 choices left for the second seat (child from one of the remaining families). The third seat is automatically filled by the remaining child.
Similarly, for the second row, there are again 3 choices for the first seat, and so on. Therefore, the number of arrangements for this case is:
Arrangements (Case 1) = 3 * 2 * 3 * 2 = 36
Case 2: Two families have siblings in one row
There are again 3 choices for which family will have siblings in the same row (A, B, or C). Without loss of generality, let's say family A has siblings in the same row.
There are now 2 choices for which child of family A sits in the first seat of the first row. The other child of family A must sit in the second seat.
The remaining child from family B can sit in the third seat. Now, for the second row, there are 2 choices for the first seat (child from either family B or C). Following the same logic as before, the number of arrangements for this case is:
Arrangements (Case 2) = 3 * 2 * 1 * 2 * 2 = 24
Total Arrangements
To get the total number of arrangements, we add the number of arrangements from both cases:
Total Arrangements = Arrangements (Case 1) + Arrangements (Case 2)
Total Arrangements = 36 + 24
Total Arrangements = 60
