+82 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?
+30 This is how I solved it. (Please tell me if it's correct.)
There are two possible arrangements of the siblings. First, there can be two pairs of siblings with each on the first or second row with one pair of siblings split between them:\( \begin{matrix} a & b & a \\ c & b & c \\ \end{matrix}\). Second, there is the arrangement where all pairs of siblings are split between rows:\( \begin{matrix} a & b & c\\ a & b & c \\ \end{matrix}\). We multiply \(2\) (the arrangements) by \(3!\) which is determined by which pairs of siblings go to \(a\), \(b\), and \(c\). Then, we multiply it by \(2^3\) which is determined by whether the first or second sibling sits in a chair for each pair of siblings. \(2(3!)(2^3)=\fbox { 96 }\)
+30 I reworked the solution with two cases, and it is actually 336.
We can have \(\begin{matrix} 6 & 4 & 1 \\ 2 & 1 & 1 \\ \end{matrix}\) for when the third seat is sibling of the first seat. There are 6 choices for the first seat. For the second seat there are 5 people left - 1 sibling of first seat = 4. Then, the third seat will be the sibling of the first seat, so 1. This leaves a pair of siblings and one person for the second row, so the only order that will work is \(\begin{matrix} s & p & s \end{matrix}\). This gives two choices from the two siblings for the second row.
Then, we will have \(\begin{matrix} 6 & 4 & 2 \\ 3 & 2 & 1 \end{matrix}\) for when the third seat is not the sibling of the first seat. 6 choices for first seat. 5 - 1 = 4 for the second seat. 4 people left - 1 sibling of first - 1 sibling of second = 2. The second row can have any order since there are no pairs of siblings left, so it is 3!.
\(6 \cdot 4 \cdot 1 \cdot 2 \cdot 1 \cdot 1 = 48\) and \(6 \cdot 4 \cdot 2 \cdot 3 \cdot 2 \cdot 1 = 288\). Summing both cases yields \(\fbox{336}\).
