+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?
+1911 We can divide this problem into two cases:
Case 1: Each row has exactly one child from each family.
Choose one child from each family for the first row: there are 3 choices
.
Arrange the remaining two children in the first row (siblings can't be together): there are 2!=2 ways.
Arrange the second row similarly: 3⋅2=6 ways total.
Case 2: One row has two children from the same family.
There are two subcases depending on the arrangement of siblings in the rows:
Subcase 2a: The first child in each row is from the same family.
Choose one pair of siblings: 3 choices
.
Arrange the siblings within the pair: 2 ways.
Arrange the remaining 4 children (2 from another pair and 2 from the third pair) in the second row: 4! ways.
Overcount: we've counted the arrangement as if sibling order matters within a pair (which it doesn't) twice (once for each sibling in the first row). So, we divide by 2!⋅2!.
Total: 2!⋅2!3⋅2⋅4!=36 ways.
Subcase 2b: The first child in the first row is NOT a sibling of the first child in the second row.
Choose one pair to have their children occupy the third seats in each row: 3 choices.
Arrange the remaining 4 children in the first row: 4! ways.
Overcount: similar to subcase 2a, we divide by 2!⋅2! to account for sibling order not mattering.
Total: 2!⋅2!3⋅4!=36 ways.
Since Cases 1 and 2 are mutually exclusive, to find the total number of arrangements, we simply add the number of arrangements from each case:
6 + 36 + 36 = 78.
This gives us a final answer of 78.
