How many choices can be made from the types of coffee beans: 4 Latte, 3 Americano, 2 Expresso, and 1 Arabica... 
How many choices can be made from the types of coffee beans: 4 Latte, 3 Americano, 2 Expresso, and 1 Arabica. If at least one type is to be chosen for coffee beans?

I have never encountered a question with this structure. I have no idea how to solve it.
There were choices given which include 120, 200, 220, and no answer. But I just can't seem to arrive anywhere near one of those choices. Please help.

UPDATE: Turns out the answer is 120, yet I still don't know how it happened.

 A: In the general case, I pick $(0/1/2/3/4)$ Latte and $(0/1/2/3)$ Americano and $(0/1/2)$ Expresso and $(0/1)$ Arabica - with the added constraint that all cannot be $0$
Using the and-or rule for counting, we can express the above as follows
$$N = \left({4 \choose 0} + {4 \choose 1} + {4 \choose 2} + {4 \choose 3} + {4 \choose 4}\right)\times \left({3 \choose 0} + {3 \choose 1} + {3 \choose 2} + {3 \choose 3} \right)\times \left({2 \choose 0} + {2 \choose 1} + {2 \choose 2}\right)\times \left({1 \choose 0} + {1 \choose 1} \right) - 1$$
The one subtracted at the end is to remove the case that all $0$
A: Type of coffee beans -
$4$ types of Latte, $3$ types of Americano, $2$ types of Espresso, $1$ type of Arabica.
Each type can be either chosen or not chosen - so $2$ possibilities. That leads to $2^4$ possibilities for Latte, $2^3$ for Americano, $2^2$ for Espresso and $2^1$ for Arabica.
So total number of combinations $ = 2^{(4+3+2+1)} = 2^{10}$. Now this will have a case where no type is chosen. That gives us the answer of $1023$.
