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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?

 Mar 15, 2024
 #1
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To solve this problem, we can break it down into steps:

Step 1: Seat the first pair of siblings in the first row. There are 3 ways to do this.

Step 2: Seat the second pair of siblings in the first row. There are 2 ways to do this (since they can't sit directly in front of their siblings).

Step 3: Seat the third pair of siblings in the first row. There is only 1 way to do this, as they must sit in the remaining seats.

Step 4: Seat the first pair of siblings in the second row. There are 2 ways to do this (since they can't sit directly in front of their siblings).

Step 5: Seat the second pair of siblings in the second row. There is only 1 way to do this, as they must sit in the remaining seats.

Step 6: Seat the third pair of siblings in the second row. There is only 1 way to do this, as they must sit in the remaining seats.

Multiplying the number of ways for each step, we get: 3 x 2 x 1 x 2 x 1 x 1 = 12

Therefore, there are 12 ways to seat three pairs of siblings from different families in two rows of three chairs following the given conditions.

 Mar 15, 2024

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