An ice cream parlor has 28 different ice cream flavors. How many different ways are there to choose 6 scoops of ice cream if at least An ice cream parlor has $28$ different ice cream flavors. How many
different ways are there to choose $6$ scoops of ice cream if at least $2$ scoops
must be chocolate?
My attempt
I used the formula $\binom {n+r-1}{r}$ and here $n=28, r=6$
$\binom {33}{6}-\binom{28}{1}$
Can anyone please explain me this
 A: Since there must be atleast 2 scoops of chocolate lets take two scoops of chocolate. Now we need to pick 4 flavors from 28. Number of ways of choosing this are are you have correctly pointed out $\binom{n+r-1}{r}$ with $n=28, r=4$, that is $\binom{31}{4}$.
A: First take $2$ scoops of chocolate. Now for the remaining $4$ scoops we have to count the number of $\geq0$ solutions of the system
$$x_1+x_2+\ldots+x_{27}+x_{28}=4\ .$$
For this count we need $4$ circles $\bigcirc$ for the scoops and $27$ separating  bars $|\>$. The result is ${31\choose4}=31\,465$.
A: No of ways of selecting scoops such that atleast 2 scoops must be chocolate
= total no of ways(S) - containing no chocolate(A) - containing exactly 1 chocolate(B)
There are 2 different answers depending on whether the arranging  order from bottom to top  matter  or not 
1)  Order matters 
S = $28^6$ since for every scoop you have 28 options
A = $27^6$ since we are excluding chocolate
B = There is exactly one chocolate which can be arranged in 6 places the rest of the places can be filled by $26^5$ ways = $6*26^5$
Answer = $S - A - B  =  28^6 - 27^6 - (6*26^5)$
2) Order doesn't matter i.e. we are only interested in different number of combinations(example ABBB,BABB,BBAB,BBBA counts as 1 case instead of 4)
The simpler way to solve this using stars and bars method  
$X_1+X_2+X_3+.......X_{28} = 6$ where $X_i$ represents the flavour of icecream
Non negative solution of above equation is given by $\binom{28+6-1}{6-1}$ =$\binom{33}{5}$
S = $\binom{33}{5}$
A = containing no chocolate = $X_1+X_2+X_3+.......X_{27} = 6$ == $\binom{27+6-1}{6-1}$ =$\binom{32}{5}$
B =  containing exactly one chocolate = 
$X_1+X_2+X_3+.......X_{27} + 1 = 6$ ($X_{28} = 1$)
=$X_1+X_2+X_3+.......X_{27} = 5$
Solutions =  $\binom{27+5-1}{5-1}$ =$\binom{31}{4}$
Answer = $S-A-B$ = $\binom{33}{5}$ - $\binom{32}{5}$ - $\binom{31}{4}$ = 4495
Alternative way:
atleast 2 chocolate
$X_1+X_2+X_3+.......X_{28} = 6$ 
let $X_{28}$ be chocolate such that $X_{28}$ >=2
Let $X_{28}$ = $Y_{28}$ +2
$X_1+X_2+X_3+.......X_{27} +Y_{28} + 2 = 6$
=$X_1+X_2+X_3+.......X_{27} +Y_{28} = 4$
Solution of above equation = $\binom{28+4-1}{4-1}$ =$\binom{31}{3}$ = 4495
