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?
+20 There are three pairs of siblings from different families to be seated in two rows of three chairs in such a way that siblings may sit next to each other in the same row, but no child may sit directly in front of their sibling.
To solve this problem, we can break it down into two cases:
Case 1: The three pairs of siblings are seated in the same row
In this case, there are 3 ways to choose which row they will sit in, and 3! ways to arrange them within that row. However, since no child may sit directly in front of their sibling, once the first pair is seated, there are only 2 possible chairs for the second child of the second pair and only 1 possible chair for the second child of the third pair. Therefore, the total number of arrangements in this case is:
3 x 3! x 2 x 1 x 1 = 36
Case 2: The three pairs of siblings are seated in different rows
In this case, there are 3 ways to choose which pair will sit in the first row, and 2 ways to choose which pair will sit in the second row. Once the pairs are chosen, there are 2! ways to arrange them within each row. Again, since no child may sit directly in front of their sibling, there are 2 possible chairs for each child in the first row and only 1 possible chair for each child in the second row. Therefore, the total number of arrangements in this case is:
3 x 2 x 2! x 2! x 2 x 2 x 1 x 1 = 96
Finally, we can add the two cases together to get the total number of arrangements:
36 + 96 = 132
Therefore, there are 132 ways to seat three pairs of siblings from different families 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.
