**Problem:** Find the different number of ways \(5\) different gifts can be presented to \(3\) children so that each child receives at least one gift.

**Solution:** There are two possible ways in which the gifts can be distributed.

**Case 1:** They are distributed as \(2,2,1\).

So first we choose the children who get \(2\) gifts each in \(^3C_2\) ways. Then we choose the gifts in \(\frac{5!}{2!.2!}\) ways.

Thus total number of ways = \(3.\frac{5!}{2!2!}= 90\) ways.

**Case 2:** They are distributed as \(3,1,1\).

So first we choose the child who gets \(3\) gifts in \(^3C_1\) ways. Then we choose the gifts in \(\frac{5!}{3!}\) ways.

Thus total number of ways = \(3.\frac{5!}{3!}= 60\) ways.

Therefore total number of ways to distribute the gifts = \(90+60\) = \(150\) ways.