<s>[INST] How can I solve this Combination with indistinguishable-objects problem? This question is not about how to solve the problem, but is about why doesn't my solution work.

A bowl has $2$ red, $2$ green, and $2$ blue marbles. How many combinations are possible if we take $3$ random marbles at a time?

I know that the answer should be, $7$. 
Coz,


*

*BBG

*BBR

*BGG

*BGR

*BRR

*GGR

*GRR


But, when I want to calculate this using a combination formula, nothing comes out which is near $7$:
$\frac{^6C_3}{2! \cdot 2! \cdot 2!}$
$ = \frac{6!}{3! \cdot (6-3)!} \cdot \frac{1}{2! \cdot 2! \cdot 2!}$
$ = \frac{6!}{3! \cdot 3!} \cdot \frac{1}{2! \cdot 2! \cdot 2!}$
$ = \frac{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{3 \cdot 2 \cdot 1 \cdot 3 \cdot 2 \cdot 1} \cdot \frac{1}{2! \cdot 2! \cdot 2!}$
$ = \frac{6 \cdot 5 \cdot 4 }{3 \cdot 2 \cdot 1} \cdot \frac{1}{2! \cdot 2! \cdot 2!}$
$ = \frac{5 \cdot 4 }{2! \cdot 2! \cdot 2!}$
$ = \frac{5}{2!} $
$ = \frac{5}{2}$
So, what am I doing wrong?
 [/INST] You can essentially break the possibilities down into 3 cases:


*

*all the colors are the same; that cannot happen so it yields 0 possibilities

*two marbles have the same color and the last one is different, as much possibilities as ordered couples of colors (6 possibilities).

*all marbles are different, 1 possibility
In the end, you get 7 possibilities.
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