There is a group of nine children, which includes two pairs of siblings (A, B) and (C, D). In how many ways can the nine children be seated in a row, so that A and B do not sit next to each other?

To solve this problem, we can use the principle of subtraction in counting.

Step 1: Calculate the total number of arrangements without any restrictions. The total number of ways to arrange 9 children in a row is 9!.

Step 2: Calculate the number of arrangements where A and B are sitting together. When A and B are together, we can treat them as one unit. So, we have 8 units to arrange, which is 8!. But A and B can switch places within their unit, so we multiply by 2!.

Step 3: Subtract the number of arrangements where A and B are together from the total number of arrangements. So, the number of ways to arrange the children so that A and B are not sitting next to each other is 9! - 2*8!.

Note: This solution assumes that all children are distinguishable and that the two pairs of siblings are distinguishable from each other.