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In how many different ways can $3$ children share $8$ identical sweets so that each child gets at least one?

I have tried this problem by listing all the possibilities and I got an answer of $21$. I also tried to solve it by using a combination formula but was unsuccessful.

So, I was wondering whether it is possible to solve this problem with such a formula because I have solved some very similar problems to this using that technique. Help would be appreciated.

Thank you :)

me_ravi_
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anonymous
  • 1,556

2 Answers2

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The sweets that have to be distributed to $3$ children are $5$ because every child has to have at least a sweet. Therefore you have to calculate the combinations with repetitions of class $5$: $$C_{3,5}= \frac{(3+5-1)!} {5!(3-1)!}=21.$$

Domenico Vuono
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0

A nice example to practise the stars & bars approach

Assume 3 children corresponds to 3 boxes(distinct) and 8 sweets(identical) corresponds to 8 stars. Now the problem is to place these 8 stars into 3 boxes. It is required that each box must have 1 star each.

So we start by first placing 1 star in each of the 3 boxes.

+-----------+
| * | * | * |
+-----------+

Now we are left with only 5 stars. Also, remember that the 3 boxes will only have 2 bars separating them.

 ***** ||

Now, all we have to do is count the total permutations of this entire pattern of 5 stars and 2 bars which is 5+2 = 7 objects, hence 7! permutations.

But, we have overcounted this way, Why? Because remember that the stars and even the bars are completely identical objects. Hence we need to remove the extra countings by dividinng it with 5! and 2!.

So the total correct ways = 7C5

i.e

21

ways to distribute 8 sweets among 3 childrens such that each gets atleast 1 sweet.

iCantC
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