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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.