TITLE: Combinatorics logic
QUESTION [7 upvotes]: Say you have a bag with one black, two white and three red balls. How
  many different ways can you pick the the balls?

Now i've tried to visualise the problem in my head, and written it down on my paper but i still can't grapple with the logic when it comes to combinatorics. 
Yes i understand that we have a total of 6 balls. Which means that for the first ball we choose we have 6 positions, now here is where my logic gets skewed. Since the second ball we pick could be white, and there are two white balls. Then the we dont have 5 places to put the second one. This is because one black, followed by two whites, and three reds is the same combination if you get what i mean. There is where my intuition gets lost. 
Any help would be greatly appreciated.

REPLY [4 votes]: Firstly, note that picking out all of the balls in turn is the same as arranging them in a row (e.g. in positions 1 to 6); I find this slightly easier to think about, so I will refer to this situation henceforth.
Method 1:
Imagine that the six (colourless) balls are already placed in a row. Now, in order to count the number of ways they may be arranged with 3 being red, 2 being white, and the remaining one being black, we can assign the 6 balls colours retrospectively.
To start with, one of the balls must be black - there are 6 choices for which ball should be black. Once this has been decided, we need two of the remaining 5 balls to be white - this can be done in ${5 \choose 2} = 10$ ways. The remaining 3 balls must then be red.
Hence, there are $6 \times 10 = 60$ ways to arrange the balls in the problem.   
Method 2:
Assuming, for now, that each ball is different/distinguishable, there are $6!$ ways of arranging the 6 balls in a row (i.e. 6 choices for the first ball, 5 for the second, etc.) I will refer to this as the simplified problem, as opposed to the actual problem.
In the actual problem, the two white balls are indistinguishable - that is, it does not matter if you swap them around, it still counts as the same arrangement. Hence, each distinct arrangement in the actual problem has been counted twice.
Similarly, the three red balls are indistinguishable; consider a single arrangement that we initially counted in our simplified problem. For each arrangement, we can hold the non-red balls fixed, and rearrange the order of the red balls, and each of these rearrangements would not count as a new arrangement in the actual problem. But in how many ways, can we rearrange these three red balls whilst holding the positions of the other colours fixed? Well there are 3 red balls, so they can be ordered/swapped in $3!$ ways. That is to say, in our simplified problem, we counted each distinct arrangement in our actual problem $3!$ times (by assuming that the red balls could be distinguished).
So, to summarise, in the simplified problem, each distinct arrangement in the actual problem was counted twice by assuming that the two white balls could be distinguished, and in fact each of these was counted  $3! = 6$ times by assuming that the three red balls could be distinguished.
Hence the number of distinct arrangement in the actual problem is $\frac{6!}{2 \cdot 3!} = 60$.