Fruit Buffet Riddle

For combinations with replacement, having n items and picking r of them can be classified as (n+r-1) choose r.
In this case, (4+3-1) choose 3 = 6 choose 3 = 20

For the choice of all the same kind of $3$ fruit slices, a customer has a choice to pick $1$ one out of the $4$ kinds, so there are $\binom{4}{1} = 4$ ways.

For the choice of $2$ kinds of fruit slices, where there are $2$ slices of the same kind and one of the other, a customer has a choice to pick $2$ one out of the $4$ kinds and then choose which one of the $2$ kinds to be the extra piece. For example, if one chooses apple and mango for the first and second piece, he/she will decide whether the third one will be of apple or mango. Thus, there are $\binom{4}{2}\binom{2}{1} = 6\times 2 = 12$ ways for this case.

Finally, for all different $3$ fruit slices, one has to pick $3$ out of the $4$ kinds, so there are $\binom{4}{3} = 4$ ways.

As a result, there are $4+12+4 = \boxed{20}$ combinations in total.

Solution 1 : We can make it into a stars and bars question. $x_1+x_2+x_3+x_4=3$ This will be $\operatorname{nCr}\left(3+4-1,3\right)$ which is $20$ . Solution 2 : generating functions. $x_1+x_2+x_3+x_4=3$ For each $x_n$ the generating function will be $\frac{1}{1-x}$ . .This will result in the generating function $\frac{1}{\left(1-x\right)^4}$ We are trying to find the coefficient of $x^3$ . Using the negative binomial theorem this will be $\operatorname{nCr}\left(-4,3\right)$ which is $\operatorname{nCr}\left(4+3-1,3\right)$ which is $20$ .