# Number of ways to choose 6 objects when they are of 11 different kinds

I have a problem that I am unable to solve. I spent too much time on this one and I think I miss something.

An ice-cream-shop has 11 flavors of ice-cream. In how many ways can one person choose 6 cornets with flavors not necessarily different?

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Does the order they are stacked on the cone matter, or is it just an unordered group of scoops in a bowl? – Ross Millikan Dec 4 '12 at 23:24
Doesn't matter... – Delcea Stefan Dec 4 '12 at 23:25
– Gerry Myerson Dec 4 '12 at 23:26

There is a trick for solving this kind of problem, or perhaps we can call it a method.

There are $11$ flavours of ice-cream. We want to choose $x_1$ of the first flavour, $x_2$ of the second flavour, and so on up to $x_{11}$ of the $11$-th flavour. The total number of cornets chosen should be $6$, so we want $$x_1+x_2+x_3+\cdots +x_{11}=6.\tag{1}$$ Naturally, quite a few of the $x_i$ will be $0$.

Every solution $(x_1,x_2,\dots,x_{11})$ in non-negative integers of Equation $(1)$ gives us a different choice of $6$ cornets, and every choice of $6$ cornets gives us a different solution of Equation $(1)$. So the number of solutions and the number of ways to get the ice-cream are the same.

I hope that finding the number of solutions of Equation $(1)$ in non-negative integers is a problem familiar to you. It is the same problem as the problem of distributing $6$ identical balls among $11$ kids. The standard technique for this problem is the Stars and Bars method, thoroughly discussed in the Wikipedia article.

The answer is given by the binomial coefficient $\dbinom{11+6-1}{6}$.

If you are not familiar with counting the number of non-negative solutions of Equation $(1)$, I recommend that you read the relevant parts of the Wikipedia article. If you experience difficulties, I can add a few lines showing how the counting is done.

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