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there are 100 pennies and 10 children

every child can get either 5, 10 or 20 pennies

How many ways to do in this case?

I assumed that n = pennies, and k = children

so if first child can get 5, 10 or 20 pennies which is n-5, n-10, or n-20

if I keep going like that, It feels like i am counting all possibilities.

I just don't know how to approach and solve this question.

ANDY
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  • Hello @ANDY, welcome to MSE. When you type mathematics in your question, please use MathJax. In this link you will find a small tutorial how to use MathJax. – Ernie060 Nov 06 '18 at 18:34
  • It rather depends on what methods you want to use. For example you could take the coefficient of $x^{100}$ in the expansion of $\left(x^5+x^{10}+x^{20}\right)^{10}$. Or you could spot you need twice as many children getting $5$ pennies as get $20$ pennies: for example there are $\frac{10!}{6!1!3!}=840$ ways $6$ children get $5$ and $1$ child gets $10$ and $3$ children get $20$. Or you could use a spreadsheet to implement a recurrence relation. – Henry Nov 06 '18 at 18:52

1 Answers1

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Consider the expansion of $(x^5 + x^{10} + x^{20})^{10}$. The result you want will be the coefficient of $x^{100}$ in the expansion.

Wolfram Alpha says it is $\boxed{4351}$.

user1952500
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