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Chapter 1: Problem 1

In how many ways can 10 (identical) dimes be distributed among five children if (a) there are no restrictions? (b) each child gets at least one dime? (c) the oldest child gets at least two dimes?

Short Answer

Expert verified
The number of ways to distribute the dimes are: (a) \({{14}\choose{4}}\) ways, (b) \({{9}\choose{4}}\) ways, and (c) \({{12}\choose{4}}\) ways.

Step by step solution

01

No restrictions

Applying the stars and bars theorem with 10 (identical) dimes and 5 children, the answer would be \({{10 + 5 - 1}\choose{5 - 1}} = {{14}\choose{4}}\) where 14 represents the total 'stars and bars', 10 being the stars (dimes) and 4 being bars (children - 1) as bars are one less than the number of children.
02

Each child gets at least one dime

In this case, each of the five children initially gets 1 dime which leaves 5 dimes left to distribute without restrictions. Hence, the total number of combinations would be \({{5 + 5 - 1}\choose{5 - 1}} = {{9}\choose{4}}\). Here, 9 represents the total 'stars and bars', 5 being the stars (remaining dimes) and 4 being bars (children - 1).
03

The oldest child gets at least two dimes

In this scenario, two dimes are given to the oldest child which leaves 8 to be distributed among all the 5 children. Hence, the number of distributions is \({{8 + 5 - 1}\choose{5 - 1}} = {{12}\choose{4}}\). Here, 12 represents the total 'stars and bars', 8 being the stars (remaining dimes) and 4 being bars (children - 1).

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