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# seating

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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 not sit next to each other in the same row?

Apr 10, 2024

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To solve this problem, let's break it down step by step:

1. **Arrange the pairs of siblings in one row**: There are 3 pairs of siblings, so we need to arrange them in one row. This can be done in $$3!$$ ways (3 factorial) because there are 3 pairs and the order matters.

2. **Arrange the pairs in the first row**: We have three pairs of siblings, and they can't sit next to each other in the same row. We can start by placing the first pair in one of the three seats, then place the second pair in one of the remaining two seats, and finally, place the third pair in the last remaining seat.

This can be done in $$3 \times 2 \times 1 = 3!$$ ways.

3. **Arrange the pairs in the second row**: After arranging the pairs in the first row, there are three pairs left to be seated in the second row. Since no siblings can sit next to each other, the first pair can be seated in any of the three available seats.

Then, the second pair can be seated in one of the remaining two seats, and the last pair will take the remaining seat. This can be done in $$3 \times 2 \times 1 = 3!$$ ways.

Now, to find the total number of arrangements, we multiply the number of arrangements for each step:

Total arrangements = (Arrangements of pairs in one row) * (Arrangements of pairs in the first row) * (Arrangements of pairs in the second row)

So, the total number of arrangements is:

$3! \times 3! \times 3! = 6 \times 6 \times 6 = 216$

Therefore, there are 216 ways to seat three pairs of siblings from different families in two rows of three chairs, if siblings may not sit next to each other in the same row.

Apr 28, 2024