+483 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?
+67 Total permutations of 6 distinct children in 6 distinct seats: 6!=720
Fix sibling pair i to sit in the same column.
The pair has two siblings s1 and s2→ 2 ways to order them vertically (top-bottom or bottom-top).
There are 3 columns → 3 possible columns for them to occupy.
The remaining 4 children can be permuted in the remaining 4 seats arbitrarily → 4!=244! = 244!=24 ways.
Therefore,
3×2×4!=3×2×24=144
2 ways to order siblings vertically → 2×2=4
3×4×2=24
6×8=48
6!−408=720−408=312
312 ways
