TITLE: Combinations: Drawing marbles from a bag
QUESTION [5 upvotes]: Problem
I'm told that a bag of marbles contains $10$ red and $5$ white marbles. Each of the red and white marbles is numbered such that they are distinct within their own groups, i.e.
$R = \{a, b, c, d, e, f, g, h, i, j\}$
$W = \{1, 2, 3, 4, 5\}$
I'm asked to calculate how many combinations there are if I am to select $4$ marbles from the bag but at least one of them must be a red marble.
Attempt
If there is at least $1$ marble, then that is $10$ choose $1$, which is just $10$. Afterwards, we are left with $14$ marbles in total and need to choose $3$ of them, which is $364$. Multiplying by $10$, I get $3,640$, but the answer key says it's supposed to be $1,360$. 
Where did I go wrong?
Edit: I tried using the counting by complement rule and got the right answer, but I'm still curious what's wrong with this approach.

REPLY [4 votes]: What's wrong with this approach? 

When you selected $1$ red marble in the first round (by 10 ways), suppose you selected $a$. Now later on, you selected 3 more marbles, let, $b,c$ and $d$. Now take a look at another case. Suppose you had selected $b$ in the first round and $a,c$ and $d$ in the second round. Don't you think these are the same cases but you have counted them repeatedly. Therefore, your answer is far greater than correct answer (Due to repetitive counting of same cases)


What to do now?

Just count the cases when there was not a single red marble selected, and subtract that number from total cases, to get the number of cases with at least one red marble.