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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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.

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