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 🙂
Best Answer
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 $5$: $$C_{3,5}= \frac{(3+5-1)!} {5!(3-1)!}=21.$$