Confused about solution to combination question Okay so here's the question:
Jason was given marbles of various colours in $4$ boxes: $3$ red marbles in the 1st box, $4$ green marbles in the 2nd box, $2$ yellow marbles in the 3rd box and $1$ black marble in the last box. How many ways can Jason choose at least $1$ marble from any of the boxes?
To solve this, I employed an approach learnt from a textbook, which is $2^{10} -1$, because I thought there are $2$ choices for each ball, picked or not, subtracting one which all are not selected. However, the answer is $4 \cdot 5 \cdot 3 \cdot 2$, its explanation being there are $4$ outcomes to three balls, which are not picked, $1$ picked, $2$ picked and so forth.
This makes me wonder how this question is different compared to another question, which is how many ways are there to select letters from a,b,c,d and e, with the solution being $2^5-1$. Why can't I use the same approach to the question above?
 A: The field of combinatorics is full of many techniques, each designed for a specific purpose.  It is important to know which tool to use in which situation; if you just grab one that looks handy you will likely get the wrong answer.
The formula $2^n-1$ counts the number of nonempty subsets of a set of size $n$ distinguishable objects.    For example, if $n=3$, and the set is $\{A,B,C\}$, then we are counting the following seven events: $$\{A\}, \{B\}, \{C\}, \{A,B\}, \{A,C\}, \{B,C\}, \{A,B,C\}$$
Since there are ten marbles altogether in the first problem, $2^{10}-1$ would be the answer if the question was how many ways are there to choose a nonempty subset of 10 distinguishable marbles.  However that is not the question; in the stated question there are boxes and colors that need to be accounted for.  Further, any two red marbles look alike.

A trick to figure out which combinatorial technique to use, is to write down a few examples of the events you are trying to count.  This should guide you in deciding whether the objects are distinguishable or not, ordered or not, and other conditions that may apply.
For the first problem, one possible outcome is (1 red, 1 green, 1 yellow, 1 black).  Another outcome is (2 red, 3 green, 1 yellow, 1 black).  Another outcome is (1 red, 4 green, 2 yellow, 1 black).  Another outcome is (1 red, 0 green, 0 yellow, 1 black).  The common pattern here is that there are four numbers, each at least 0, but not all zero.  The first number is in the range 0-3, the second in the range 0-4, the third in the range 0-2, and the last in the range 0-1.  Hence there are $4\times 5\times 3\times 2=120$ events we are counting, but one of these is (0,0,0,0), so we want only 119.
