Use of a null option in a combination with repetition problem 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!

Without using a null, the 'standard calculation' is very annoying. The standard idea would be to see how many ways no scoops can be taken, plss how many ways 1 can be taken, plus 2, ... plus 19. Using a null simplifies all that to one step, as having 1 null just means that you are looking at 18 scoops. 2 nulls means 17 scoops, etc.
It's more troublesome that you just took ${ n + r - 1 \choose n - 1}$.With 20 and 6, you are literally saying how many ways exactly 20 scoops can be chosen from 6 flavors, which is not at all what the question asks. But perhaps you already considered combinatorial identities?
