1. ## combinatorics question

Hello everybody,
This is probably a simple combinatorics question.

- We do have $A$ red balls, $B$ green balls and $C$ yellow balls ( $A, B, C$ is their count, so they are totall $A+B+C$ in count). If we combine them together into $n$-pairs how many are their unique combinations?

I tried for the combination into $2$-pairs and i solved it for special cases (example $A=3, B=1, C=1$).
$\left(\begin{array}{cc}3\\2\end{array}\right)+
\left(\begin{array}{cc}A-1\\2\end{array}\right)+
\left(\begin{array}{cc}B-1\\2\end{array}\right)+
\left(\begin{array}{cc}C-1\\2\end{array}\right)=4$

However if i try to generalize it, I fail. I know that I am confused and this is probably wrong proccess to the problem.

Thank you all for your time.

2. Originally Posted by gdmath
We do have $A$ red balls, $B$ green balls and $C$ yellow balls ( $A, B, C$ is their count, so they are totall $A+B+C$ in count). If we combine them together into $n$-pairs how many are their unique combinations?
That is a somewhat vague description.
By 'pairs' do you mean ordered pairs, simple sets of two each, or something else altogether?
It is simply not clear what a "3-pair" could be.
Can you clear this up? Give an example: say $A=3,~B=2,~C=1$.

3. I suppose the question asks for non-ordered pairs and balls in each colour are non-identical.

RED-RED pairs: ${{A}\choose{2}}$

GREEN-GREEN pairs: ${{B}\choose{2}}$

YELLOW-YELLOW pairs: ${{C}\choose{2}}$

RED-GREEN pairs: ${{A}\choose{1}}{{B}\choose{1}}$

GREEN-YELLOW pairs: ${{B}\choose{1}}{{C}\choose{1}}$

YELLOW-RED pairs: ${{C}\choose{1}}{{A}\choose{1}}$

Add these to get the total number of unique pairs. This is actually the total number of pairs each of which are different.

Originally Posted by gdmath
I tried for the combination into $2$-pairs and i solved it for special cases (example $A=3, B=1, C=1$).
$\left(\begin{array}{cc}3\\2\end{array}\right)+
\left(\begin{array}{cc}A-1\\2\end{array}\right)+
\left(\begin{array}{cc}B-1\\2\end{array}\right)+
\left(\begin{array}{cc}C-1\\2\end{array}\right)=4$
If $B,C$ are 1, the above expression becomes meaningless.

EDIT: By the way, after working it out it seems to me here PAIR means any number of balls taken together; is that right?

4. First of all thank you for your time and your response.

Let me give you an example:

We have:
- 4 Balls of red (A=4)
- 3 Balls of green (B=3)
- 5 Balls of yellow (C=5)

If we pick up 3 (n=3) balls each time, how many possible combinations (unique combinations) can have?
We do not care about the distribution of combinations. We only care if we can count the unique combinations.

for this example, by observation we do have 10 unique combinations:
(we do not care for the order)

Red - Red - Green
Red - Green - Green
Yellow - Green - Green
Yellow - Yellow - Green
Red - Red - Yellow
Red - Yellow - Yellow
Red- Green- Yellow
Red-Red-Red
Green-Green-Green
Yellow - Yellow-Yellow

Is it possible to use a generalized formula with A, B, C and (mostly) n?

THANK YOU A LOT

5. Originally Posted by gdmath
We have:
- 4 Balls of red (A=4)
- 3 Balls of green (B=3)
- 5 Balls of yellow (C=5)
If we pick up 3 (n=3) balls each time, how many possible combinations (unique combinations) can have?
We do not care about the distribution of combinations. We only care if we can count the unique combinations. for this example, by observation we do have 7 unique combinations:
(we do not care for the order)
Thank you for the nice clarification.
In this particular example, the answer is simple because the number of selections, $n=3$, is not more than the minimum of color calls, the greens.
The answer is $\dbinom{3+3-1}{3}=10$.

In the case $n>\min\{A,B,C\}$ the process is more complicated.

6. In the case $n>\min\{A,B,C\}$ the process is more complicated.
Let me assume that there is no a generalized formula of doing that.
Furthermore the process depends on the arithetic data and might change on specific ocussions?

THANK YOU A LOT.

7. Originally Posted by gdmath
Let me assume that there is no a generalized formula of doing that. Furthermore the process depends on the arithetic data and might change on specific ocussions?
Well if $n\le\min\{A,B,C\}$ then there general formula:
$\dbinom{n+2}{n}$.