Alice wants to make a fruit salad using 10 pieces of fruit. She can buy three types of fruits at the supermarket: Apples (A), Bananas (B), and Cranberries (C). She can take as many of each type of fruit as she wants but she can only have 10 in the end. One possible fruit salad can be made of 5 Apples, 3 Bananas, and 2 Cranberries. How many types of fruit salads can Alice make

One fruit salad differs from the other only in the amount of pieces of certain fruit put in it. In order to easier denote fruit pieces we introduce these notations:

A-how many apples are put into the salad;

B-how many bananas are put into the salad;

C-how many cranberries are put into the salad.

Since she can freely choose the number of pieces of each fruit, we have these conditions for the variables A, B and C:

[tex]0\leq A\ ,\ 0\leq B\ ,\ 0\leq C[/tex] (she cannot choose a negative number of pieces)
[tex]\ A\ ,\ B\ ,\ C\ \leq 10[/tex] (because she can get the total of 10 pieces of fruit)

Another condition for forming the salad is that the salad must consist of exactly 10 pieces of fruit, hence we have this equation to solve:

[tex]A+B+C=10[/tex]

but we must obtain the non-negative integer solutions of this equation.

That is equivalent to calculating the number of r-combinations of the multi-set S with objects of k different types with infinite repetition numbers.

The formula for obtaining the number of such r-combinations is:

[tex]{r+k-1\choose r}={r+k-1\choose k-1}[/tex]

We have that [tex]k=3[/tex] and that [tex]r=10[/tex] and we can observe the repetition number as infinite since she can create a fruit salad with only one piece of fruit and the repetition number in such cases is the maximum 10. Finally, we have that the total number of fruit salads equals:

[tex]{10+3-1\choose 10}={12\choose 10}=\frac{12!}{10!\cdot (12-10)!}=\frac{12\cdot 11\cdot 10!}{10!\cdot 2!}=\frac{132}{2}=66[/tex] .