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Test of Mathematics Solution Subjective 48 - The Gifts Distribution

Test of Mathematics at the 10+2 Level

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.


Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

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.

 

Test of Mathematics at the 10+2 Level

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.


Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

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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