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?
This question has been asked a lot ot times recently.
there are probably a number of different answers given.
I let the children be A1 A2 B1 B2 C1 C2
If there were no restrictions there would be 6! ways to seat them
How many ways can they NOT be seated.
We cannot have all the pairs under each other so that would be
6 places to put A1 then 1 place to put A2
4 places to put B1 then 1 place to put B2
2 places to put C1 then 1 place to put C2
6*4*2 = 48
We cannot have one pair vertically alligned with the others not so
Say the As are vertially alligned but not B or C
6 places to put A1, 1 place to put A2
4 places to put B1, 2 places to put B2
2 places to put C1 and 1 place to put C2
6*1*4*2*2*1 = 96 ways
BUT ic could have been A or B or C that are vertically aligns so we have to multiply by 3
So I think the answer is
6! - (48+288) = 6! - 336 = 720 - 336 = 384 ways
Take anyone of the 6 kids:
He/she has 6 choices. The sibling has 5 - 1 = 4 choices.
The 4th kid has the remaining 4 choices. The sibling has 3 - 1=2 choices
The last pair has the remaining 2 choices.
Total seating =6 x 4 x 4 x 2 x 2 =384 ways of seating the 3 pairs of siblings.