I am really stuck on this
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?
+40 Let this be the rows of chairs:
[] [] []
[] [] []
Let their be siblings \(x\) and \(x\), \(y\) and \(y\), and \(z\) and \(z\). Beginning with placing the \(x\) siblings, there are 6 options for the first sibling obviously, and 4 options for the second sibling, giving us 24 ways of placing them (6*4):
[x1] [] []
[x2 not allowed] [] []
Let's say we go with this arrangement:
[x1] [] []
[] [x2] []
We must place one of the y siblings in the rightmost column, or else the z siblings will be front and behind each other (there are two ways to do this):
[x1] [] []
[] [x2] [y]
Then there are two ways of placing the other y sibling (row 1 column 2 or row 2 column 1), giving us 4 ways of placing the y siblings (2*2), but then we could also swap it so that sibling y2 could be in the right most column and y1 is in the other spots, givings us 8 ways of placing the y siblings:
[x] [] []
[y2] [x] [y1]
And then the z siblings go in automatically, with 2 ways, as we can swap z1 and z2. this gives us \(24*8*2=\boxed{384}\) ways. Sorry if my explanations a bit long, this problem I found hard to explain without going super in depth.
