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?

blackpanther Dec 16, 2023

#1**0 **

Here's how to solve the problem for three pairs of siblings:

Approach through cases: Again, we can split the problem into cases based on the first pair of siblings:

Case 1: Pair sits together in either row:

Choose which row the first pair sits in (2 options).

Arrange the pair within the chosen row (2 options).

Arrange the remaining eight children in the other row (8! options).

Case 2: Pair doesn't sit together:

Choose one sibling from the pair to sit in the first row (2 options).

Arrange the chosen sibling and the remaining three children in the first row (4! options).

Place the other sibling from the pair in the second row (1 option).

Arrange the remaining seven children in the second row (7! options).

Combine case results: Multiply the options for each case:

Total arrangements = (Case 1 options) + (Case 2 options) = (2 * 2 * 40320) + (2 * 4! * 1 * 5040) = 161280 + 28800 = 189680

Therefore, there are 189680 ways to seat the four pairs of siblings with the given restrictions.

Note: Calculating 8! and 7! directly can be cumbersome. You can simplify the calculation by observing patterns:

8! = 8 * 7!

7! = 7 * 6!

6! = 6 * 5!

Therefore, 8! can be expressed as 8 * 5! * 7 * 6 and 7! as 7 * 5! * 6. Substituting these back into the final equation:

Total arrangements = (2 * 2 * 8 * 5! * 7 * 6) + (2 * 4! * 1 * 7 * 5! * 6) = (16 * 5! * 42) + (14 * 5! * 42) = 42 * 5! * (16 + 14) = 42 * 5! * 30 = 189680

The answer is 189680.

BuiIderBoi Dec 17, 2023