Why sum three combinations for at most 2 mochas in 10 coffees

[Superdemongob]

This is not a homework problem but in fact a solved example question.
The full question text is below with answers and my reasoning for them. Could someone please explain the reasoning behind part (c).

A coffee shop sells 5 types of coffee (latte, mocha, espresso, cappuccino and iced coffee). All coffee of the same type is indistinguishable. We are buying 10 coffees in total.

How many ways are there to buy 10 coffees if:

(a) there are no restrictions

C(10+5-1, 10)​

The reasoning is that given no restrictions, this is a combination with repetition.​

(b) you buy at least 2 iced coffees.

C(8+5-1, 8)​

The reasoning is that since 2 coffees are decided, its a combination with repetition for the remaining 8.​

(c) you can buy at most 2 mochas.

C(13,3) + C(12,3) + C(11,3)​

Here is where I have no idea how they got these numbers.​

Any help is greatly appreciated.

 

[justsomeguy]

Personally I'd do it as
10 non-mocha cups (10 from the set of 4).
9 non-mocha cups and a mocha (9 from the set of 4, 1 from the set of 1) .
8 non-mocha cups and two mochas (8 from the set of 4, 2 from the set of 1).

Note that C(10+5-1, 10) == C(10+5-1, 5-1).

It looks like you're more comfortable with the first type but the answer for c is using the 2nd type; e.g. C(13,3) = C(10+4-1, 4-1) = C(10+4-1, 10)

Does that help you any?

 

[mfb]

Those parts are:
You buy 0 mochas + you buy 1 mocha + you buy 2 mochas, with the same formula as in (a) (if you replace (n choose k) by (n choose n-k)).