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  • How many different ways may 12 identical candies be distributed among four children if each child receives at least one candy but no more than four?  <str

How many different ways may 12 identical candies be distributed among four children if each child receives at least one candy but no more than four?

 

Option: 1

21


Option: 2

31


Option: 3

41


Option: 4

51


Answers (1)

best_answer

Given that,

There are 12 identical candies which are distributed among 4 children.

Each child receives at least one candy but no more than four.

Four children each have one candy.

The remaining 8 candies(12 initially dispersed) can be distributed among these with the constraint that no one receives more than 3 (since this would increase the total sum to more than 4) but can receive 0 (as they already have a candy).

The possible number of cases is,

Case 1: 

The distribution of candies will be as follows: 3 + 3 + 2 + 0

Thus, the number of ways the candies are distributed is given by,

\begin{aligned} & \frac{4 !}{2 !}=\frac{4 \times 3 \times 2}{2} \\ & \frac{4 !}{2 !}=12 \end{aligned}

Case 2: 

The distribution of candies will be as follows: 3 + 3 + 1 + 1

Thus, the number of ways the candies are distributed is given by

\begin{aligned} &\frac{4 !}{2 ! 2 !}=\frac{4 \times 3 \times 2}{2 \times 2}\\ &\frac{4 !}{2 ! 2 !}=6 \end{aligned}

,Case 3: 

The distribution of candies will be as follows: 3 + 2 + 2 + 1

Thus, the number of ways the candies distributed is given by,

\begin{aligned} & \frac{4 !}{2 !}=\frac{4 \times 3 \times 2}{2} \\ & \frac{4 !}{2 !}=12 \end{aligned}

Case 4: 

The distribution of candies will be as follows: 2 + 2 + 2 + 2


Thus, the number of ways the candies are distributed is given by,

\frac{4 !}{4 !}=1

Therefore, the total number of possible ways is 12 + 6 + 12 + 1 = 31 ways.

 

 

 

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