Ok, so I am working on a combinatorics problem involving combination with repetition. The problem comes off a past test that was put up for study. Here it is:
An ice cream parlor sells six flavors of ice cream: vanilla, chocolate, strawberry, cookies and cream, mint chocolate chip, and chocolate chip cookie dough. How many combinations of fewer than 20 scoops are there? (Note: two combinations count as distinct if they differ in the number of scoops of at least one flavor of ice cream.)
Now I get the correct answer $\binom{25}{6}$, but the way they arrive at the answer is different and apparently important. I just plug in 20 combinations of 6 flavors into $\binom{n+r-1}{r}=\binom{n+r-1}{n-1}$. The answer given makes use of a "null flavor" to be used in the calculation. I can't figure out for the life of me why, could someone explain this to me?
Answer:
This is a slight variation on the standard combinations with repetition problem. The difference here is that we are not trying to buy exactly 19 scoops of ice cream, but 19 or fewer scoops. We can solve this problem by introducing a 7 th flavor, called “noflavor” ice cream. Now, imagine trying to buy exactly 19 scoops of ice cream from the 7 possible flavors (the six listed an “no-flavor”). Any combination with only 10 real scoops would be assigned 9 “no-flavor” scoops, for example. There is a one-to-one correspondence between each possible combination with 19 or fewer scoops from 6 flavors as there are to 19 “scoops” from 7 flavors. Thus, using the formula for combination with repetition with 19 items from 7 types, we find the number of ways to buy the scoops is $\binom{19+7-1}{19}=\binom{25}{19}=\binom{25}{6}$. (Grading – 4 pts for mentioning the idea of an extra flavor, 4 pts for attempting to apply the correct formula, 2 pts for getting the correct answer. If a sum is given instead of a closed form, give 6 points out of 10.)
Any assistance would be greatly appreciated!