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?
+195 We can solve this problem by using the Principle of Inclusion-Exclusion (PIE). We first count the total number of ways to seat the six children without any restrictions.
There are 6 choices for the first seat, 5 choices for the second seat (since one child has already been seated), and so on, for a total of 6! = 720 possible arrangements.
Next, we count the number of arrangements where at least one pair of siblings sit next to each other in the same row. We can choose one of the three pairs of siblings to sit together, and then there are 2 ways to arrange these siblings within the pair. There are 4 choices for the first seat in the row containing the pair, and then 3 choices for the second seat (since one child is already seated), for a total of 3 * 2 * 4 * 3 * 2 = 144 arrangements. However, we have overcounted the arrangements where two pairs of siblings sit together, so we need to subtract them off.
There are 3 ways to choose which two pairs of siblings sit together, and then there are 2 ways to arrange each pair within their respective row. There are 2 choices for the first seat in the row containing the first pair, and then 1 choice for the second seat (since the other child in that pair is already seated), and 2 choices for the first seat in the row containing the second pair, and then 1 choice for the second seat (since the other child in that pair is already seated), for a total of 3 * 2 * 2 * 1 * 2 * 1 = 24 arrangements.
However, we have subtracted too much, since there is one arrangement where all three pairs of siblings sit together in the same row that we have subtracted twice. There are 3! = 6 ways to arrange the three pairs within the row, and then 3! = 6 ways to arrange the rows, for a total of 6 * 6 = 36 arrangements.
Therefore, the number of arrangements where at least one pair of siblings sit together in the same row is 144 - 24 + 36 = 156.
Finally, we subtract this number from the total number of arrangements to get the number of arrangements where no pair of siblings sit together in the same row:
720 - 156 = 564.
Therefore, there are 564 ways to seat the 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.
A1 has 6 options to choose from. A2 (his/her sibling) has 5 - 1=4 options to choose from. B1 has the remaining 4 options to choose from. B2 has 3 - 1 = 2 options to choose from. The remaining 2 chairs are for the last pair C1 and C2.
So, in total, we have: 6 x 4 x 4 x 2 x 2 ==384 ways of seating the 3 pairs of children
