A shop has $4$ distinct flavors of ice-cream. One can purchase any number of scoops of any flavor. The order in which the scoops are purchased is inconsequential.If one wants to purchase $3$ scoops of ice-cream, in how many ways can one make that purchase?

The problem asks for the number of combinations with repetition allowed.We have $n=4$ distinct flavors (categories) and we want to choose $r=3$ scoops (items).The formula for combinations with repetition (often solved using the "Stars and Bars" method) is:
$$\text{Number of ways}=\left(\genfrac{}{}{0ex}{}{n+r-1}{r}\right)$$
Substitute $n=4$ and $r=3$:
$$\text{Number of ways}=\left(\genfrac{}{}{0ex}{}{4+3-1}{3}\right)=\left(\genfrac{}{}{0ex}{}{6}{3}\right)$$
Now, evaluate the binomial coefficient:
$$\left(\genfrac{}{}{0ex}{}{6}{3}\right)=\frac{6!}{3!(6-3)!}=\frac{6\times 5\times 4}{3\times 2\times 1}=\frac{120}{6}=20$$
There are $20$ different ways to purchase the $3$ scoops.