Choosing toppings for pizza Problem: Ordering a "deluxe" pizza means you have four choices from 15 available toppings. How many combinations are possible if toppings cannot be repeated? If they can be repeated? Assume the order in which the toppings are selected does not matter.
If the toppings cannot be repeated, then we have $C_4^{15}$ choices.
I am having difficulty with the toppings can be repeated problem.  It feels like this isn't solvable with the stars and bars method because we only care about "4" of the "15" toppings and not all 15.
What I think: Use "14" bars to then have "15" categories, each responding to a possible topping.  Have 4 stars to represent choosing a topping from that category.  Then we should get $C^{18}_{4}$?
 A: This may not be the most efficient way of doing this, but you can consider it as separate cases.


*

*If all toppings are distinct, then you have $C_4^{15}$ combinations.

*If there are three distinct toppings, you have $3 \cdot C_3^{15}$ combinations (because we have $C_3^{15}$ choices for toppings and then $3$ choices for which of those three toppings is doubled).

*If there are two distinct toppings, you have $3 \cdot C_2^{15}$ combinations (because there are $C_2^{15}$ choices for topping and $3$ possibilities: either both toppings doubled, the first is tripled, or the second is tripled).

*If there is only one distinct topping, you have $15$ possibilities.
Add these up to get your total.

Let's enumerate them with $5$ options instead of $15$ for illustrative purposes
Case 1:
ABCD
ABCE
ABDE
ACDE
BCDE

Case 2:
AABC  ABBC  ABCC
AABD  ABBD  ACDD
AABE  ABBE  ABEE
AACD  ACCD  ACDD
AACE  ACCE  ACEE
AADE  ADDE  ADEE
BBCD  BCCD  BCDD
BBCE  BCCE  BCEE
BBDE  BDDE  BDEE
CCDE  CDDE  CDEE

Case 3:
AABB
AACC  BBCC
AADD  BBDD  CCDD
AAEE  BBEE  CCEE  DDEE

and
      BBBA  CCCA  DDDA  EEEA
AAAB        CCCB  DDDB  EEEB
AAAC  BBBC        DDDC  EEEC
AAAD  BBBD  CCCD        EEED
AAAE  BBBE  CCCE  DDDE

Case 4:
AAAA
BBBB
CCCC
DDDD
EEEE

A: if they can be repeated, it can be stars and bars - you have 15 things to choose from, but you only want to choose 4 - suppose you chose only 4 onion
****||||||||||||||
is a basic representation of 4 going into 15 slots
4 stars and 14 bars
so it is $\binom{18}{4} $
if they can't be repeated, then it is 4 from 15 where there is one of each, without replacement and without order, so it is $\binom{15}{4}$
A: You are correct if the toppings cannot be repeated.  If they can, presumably order does not matter, but you need to specify if which one is repeated does.  Is AABC different from ABBC because in the first case you get twice as much A as C?  
Assuming it matters which are repeated, you can have four different toppings, two of one and one each of two others, three of one and one other, or ... For two of one and one each of two others you have $15$ choices for the repeated one and $14 \choose 2$ for the others.  Do this for each pattern and you are done.
