How can I solve this Combination with indistinguishable-objects problem?

Question

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,

1. BBG
2. BBR
3. BGG
4. BGR
5. BRR
6. GGR
7. 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?

Answer 1

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.

Answer 2

"The question is not about how to solve the problem, but was about why doesn't my solution work."

Continuing from an earlier point I made in the comments, it is much more important to rely on techniques than on formulas at this early level. Any formula you encounter can be replicated on the spot whenever you need it using the techniques you have learned, including combinations, permutations, stars and bars, combinations with repitition, etc...

I mentioned elsewhere how rule of sum and rule of product are essential introductory counting techniques.

Another very useful counting technique is sometimes referred to as the following:

The Shepherd's Principle: When counting objects or arrangements, if your current count is $n$ but you have overcounted each object/arrangement the exact same number of times, say $k$ times each, then you may correct your count by dividing: $\frac{n}{k}$

The name comes from that if you know that every sheep has four legs each and you wish to count the number of sheep in your field and can see every leg but cannot distinguish one body from another or just don't have a clear view of the bodies, you may count how many legs you see and divide by four.

The shepherd's principle explains where most of the division comes from in formulas and techniques used in counting. It can be seen for example in the formula for combinations, $\binom{n}{r}=\frac{n!}{r!(n-r)!}$, and is one among many ways of proving that formula to be correct. Related to your example, the number of ways to arrange all six of the marbles in a line where order matters would be $\frac{6!}{2!2!2!}$, recognized by temporarily assuming the balls are all different colors making $6!$ arrangements and recognizing that each arrangement was overcounted a total of $2!2!2!$ times each.

Why didn't it work for you here? Because many of the arrangements you were counting were not counted $2!2!2!$ times each and some arrangements were overcounted more times than others.

Among the $\binom{6}{3}=20$ ways in which we may select three marbles if we were to temporarily assume they were all distinct, it breaks into cases. Eight of which correspond to all of the colors actually being different. The remaining twelve ways correspond to two of the colors actually matching and the remaining color being different. In the case that all three balls were actually different colors, this was accidentally counted $2!2!2!=8$ times but should only have been counted once. In the case that two of the balls were the same actual color, these were only accidentally counted twice each, not eight times each, yielding a corrected count for this case as being $\frac{12}{2}=6$.

This would give the corrected count as $\frac{8}{2!2!2!}+\frac{12}{2}=1+6=7$, giving the correct answer.

I personally find this to be a rather unhelpful technique to be used for this problem, but it can be forced to work, especially since in order to have counted the number of arrangements which fell into the specific cases it requires as much effort as going through the problem a more direct way to begin with essentially making the application of the shepherd's principle unnecessary extra work.

Answer 3

use generating function to count :

you can select either $0$ ball ; $1$ ball ; $2$ ball of same colour

so $f(x)= (x^0+x+x^2)(x^0+x+x^2)(x^0+x+x^2)=(1+x+x^2)^3$

so, to find number of ways to pick $3$ balls out of the bowl, just find coefficient of $x^3$ in expansion of $f(x)$

now , coefficient of $x^3$ in expansion of $f(x)=(1+x)^3+3.(1+x)^2.x^2+3(1+x)x^4+x^6$ is $7$ so there will be 7 combinations of drawing a ball from the bowl.