Number of ways to choose $4$ objects out of $6$ groups with $3$ members each Suppose a box contains 18 balls number 1-6, three balls with each number. When 4 balls are drawn without replacement, how many outcome are possible?
I took $6\choose4$ ways of picking the balls and then took away the possibilities where 4 of the same ball were chosen as this is impossible (can be done in 6 ways)
This gives me a final answer of; $${6\choose4} - 6$$
My question is;
Is this method/answer correct and also is there a general way to solve this type of problem?
Note; This problem is from a book called 'An introduction to Combinatorics and Graph Theory' and I think it comes under the umbrella of multisets and upper bounds.
 A: Let $x_k$ denote the number of balls numbered $k$ that are selected.  If four of the $18$ balls are selected, then 
$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 4 \tag{1}$$
where $x_k \leq 3$ for $1 \leq k \leq 6$.   Equation 1 is an equation in the non-negative integers.  A particular selection corresponds to a choice of where to place five addition signs in a row of four ones.  For instance,
$$1 + 1 + + 1 + + 1$$
corresponds to $x_1 = 1$, $x_2 = 1$, $x_3 = 0$, $x_4 = 1$, $x_5 = 0$, and $x_6 = 1$.  Hence, if there were no restrictions, the number of solutions of equation 1 would be the number of ways five addition signs could be inserted into a row of four ones, which is
$$\binom{4 + 5}{5} = \binom{9}{5}$$
since we must choose which five of the nine symbols (four ones and five addition signs) will be addition signs.  However, six of these solutions violate the restriction that $x_k \leq 3$ for $1 \leq k \leq 6$, namely those in which $x_k = 4$ for some $k$ satisfying $1 \leq k \leq 6$.  Hence, the number of ways to select four balls from $18$ balls consisting of three balls apiece numbered from 1 - 6 is $$\binom{9}{5} - 6$$
A: $\binom{6}{4}-6$ counts ways to select four colours minus ways to select one colour.
That's not right at all.

You wish to count the distinct numbers and colours that may be selected.
As to numbers, you can select from restricted partitions of 4. Specifically: $(3:1), (2:2), (2:1:1), (1:1:1:1)$
That is you may select either: a triple-and-single, two-doubles, a double-and-two-singles, or four-singles.
But how many ways can you select colours for each of these alternatives?   (There's a lot more than nine)
An alternative approach is to use the stars-and-bars method to count ways to put four identical balls into six bins such that no more than three go into the same bin.
$$\mathop{\underline{\binom{6}{1,1,4}}}_\textrm{#triple-&-single}+\mathop{\underline{binom{6}{2,4}}}_\textrm{#two-doubles}+\mathop{\underline{\binom{6}{1,2,3}}}}_\textrm{#double-&-two-singles}+\mathop{\underline{\binom{6}{4,2}}}_\textrm{#four-singles} = \underline{\binom{4+6-1}{6-1}-6}$$
