First, arrange the five older children. There are 5! ways to do this.

Next, consider the gaps between and around these five children where the three youngest can be placed. There are 6 such gaps.

Choose 3 out of these 6 gaps to place the youngest children. This can be done in \(\binom{6}{3} = 20\) ways.

Arrange the three youngest children in these chosen gaps. There are 3! ways to arrange them.

Thus, the total number of ways is \(5! \times \binom{6}{3} \times 3! = 120 \times 20 \times 6 = 14400\).