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This is a Test of Mathematics Solution Subjective 48 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

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

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.

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