Try this beautiful Problem on Combinatorics from integer based on chocolates from PRMO -2018

## Chocolates Problem – PRMO 2018- Problem 28

Let N be the number of ways of distributing 8 chocolates of different brands among 3 children such that each child gets at least one chocolate, and no two children get the same number of chocolates. Find the sum of the digits of $\mathrm{N}$.

,

- \(28\)
- \(90\)
- \(24\)
- \(16\)
- \(27\)

**Key Concepts**

Combination

Combinatorics

Probability

## Suggested Book | Source | Answer

#### Suggested Reading

Pre College Mathematics

#### Source of the problem

Prmo-2018, Problem-28

#### Check the answer here, but try the problem first

\(24\)

## Try with Hints

#### First Hint

we have to distribute \(8\) chocolates among \(3\) childrens and the condition is Eight chocolets will be different brands that each child gets at least one chocolate, and no two children get the same number of chocolates. Therefore thr chocolates distributions will be two cases as shown below…..

Now can you finish the problem?

#### Second Hint

case 1:$(5,2,1)$

Out of \(8\) chocolates one of the boys can get \(5\) chocolates .So \(5\) chocolates can be choosen from \(8\) chocolates in \( 8 \choose 5\) ways.

Therefore remaining chocolates are \(3\) . Out of \(3\) chocolates another one of the boys can get \(2\) chocolates .So \(2\) chocolates can be choosen from \(3\) chocolates in \( 3 \choose 2\) ways.

Therefore remaining chocolates are \(1\) . Out of \(1\) chocolates another one of the boys can get \(1\) chocolates .So \(1\) chocolates can be choosen from \(1\) chocolates in \( 1 \choose 1\) ways.

Therefore number of ways for first case will be \( 8 \choose 5\) \( \times\) \( 3 \choose 2\) \( \times\) \( 1 \choose 1\)\(\times\) $3!$=$\frac{8}{2!.5!.1!}$$\times 3$

Case 2:$(4,3,1)$

Out of \(8\) chocolates one of the boys can get \(4\) chocolates .So \(4\) chocolates can be choosen from \(8\) chocolates in \( 8 \choose 4\) ways.

Therefore remaining chocolates are \(4\) . Out of \(4\) chocolates another one of the boys can get \(3\) chocolates .So \(3\) chocolates can be choosen from \(4\) chocolates in \( 4 \choose 3\) ways.

Therefore remaining chocolates are \(1\) . Out of \(1\) chocolates another one of the boys can get \(1\) chocolates .So \(1\) chocolates can be choosen from \(1\) chocolates in \( 1 \choose 1\) ways.

Therefore number of ways for first case will be \( 8 \choose 4\) \( \times\) \( 4 \choose 3\) \( \times\) \( 1 \choose 1\)\(\times\) $3!$=$\frac{8}{4!.3!.1!}$$\times 3$

Can you finish the problem…?

#### Third Hint

Therefore require number of ways =$\frac{8}{2!.5!.1!}$$\times 3$+$\frac{8}{4!.3!.1!}$$\times 3$=$24$

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