# Thread: Set of points, 3 objects, get all possible combinations

1. ## Set of points, 3 objects, get all possible combinations

Hi,

What formula would I use to solve something like this:

If I have 12 possible positions and a certain number of objects e.g. 3 and need to figure out all the combinations for the 3 objects across the 12 positions, how would I go about this?

If you someone could direct me to the correct formula or a similar problem/solution I would be very grateful.

Thanks.

2. ## Re: Set of points, 3 objects, get all possible combinations

Originally Posted by kerrymaid
Hi,

What formula would I use to solve something like this:

If I have 12 possible positions and a certain number of objects e.g. 3 and need to figure out all the combinations for the 3 objects across the 12 positions, how would I go about this?

If you someone could direct me to the correct formula or a similar problem/solution I would be very grateful.

Thanks.
Does the permutation of the objects within the three positions matter? If so, then you multiply the number of positions that the first object could be in (12), by the number of positions the second object could be in (11), and so on, until you've run out of objects. If differing permutations within the three used positions aren't counted as unique, then you must divide it by n! in order to erase the non-unique permutations, where n is the number of objects being placed.

3. ## Re: Set of points, 3 objects, get all possible combinations

Thanks for the reply: so as a quick sanity check:
If I have 12 possible positions and 3 objects, there should be (12x11x10)/(1x2x3) = 220 combinations?

There should be only one object in a given position at any time...However I also need to eliminate scenarios such as the following:

object 1: position 1, object 2: position 5, object 3: position 12
object 1: position 12, object 2: position 1, object 3: position 5

For the purposes that I need this for that would be considered the same combination as there are still three objects (it doesnt matter which ones) in the same positions.

4. ## Re: Set of points, 3 objects, get all possible combinations

Is the above correct?

5. ## Re: Set of points, 3 objects, get all possible combinations

I;ve been doing some reading and maybe I have this wrong...I think the formula should be n!/(n-r)! or 12!/(12-3)! which gives me 1320 combinations...but I;m not sure if that accounts for the repetition that I described above....

6. ## Re: Set of points, 3 objects, get all possible combinations

Since order doesn't matter, the answer is "12 choose 3" = 12! / (9! x 3!).
12! is the number of permutations of the 12 positions.
For each permutation of the first three, there are 9! permutations of the remaining positions. So you need to divide by 9! to eliminate over counting.
And since the order of the first three doesn't matter, you need to divide by 3! to come up with combinations.

Same result as Feryll's description, just stated a bit differently.

7. ## Re: Set of points, 3 objects, get all possible combinations

Hi flatiron,

Thanks for the answer...Probably a bit of a daft question but if I'm following the above correctly does that mean there are the same possible amount of combinations for 2 objects as 10 objects:

No Positions: 12, Objects: 2: "12 choose 2" - 12! / (10! * 2!)
No Positions: 12, Objects: 10: "12 choose 10" - 12! / (2! * 10!)

Both options give me 66 combinations.

That seems off...

8. ## Re: Set of points, 3 objects, get all possible combinations

Originally Posted by kerrymaid
Probably a bit of a daft question but if I'm following the above correctly does that mean there are the same possible amount of combinations for 2 objects as 10 objects:
No Positions: 12, Objects: 2: "12 choose 2" - 12! / (10! * 2!)
No Positions: 12, Objects: 10: "12 choose 10" - 12! / (2! * 10!)
Both options give me 66 combinations.
That seems off...
That may be, because the question seems a bit off.
Lets make it simple.
Suppose we have a board with twelve peg holes in a row.
We have three blue pegs.
There are $\binom{12}{3}=220$ ways to put those pegs into the peg holes.

If we have two blue pegs.
There are $\binom{12}{2}=66$ ways to put those pegs into the peg holes.

On the other hand.
If we have three pegs, one red, one white, and one blue.
There are $12\cdot 11\cdot 10=1320$ ways to put those pegs into the peg holes.

If we have two pegs, one red and one blue.
There are $12\cdot 11=132$ ways to put those pegs into the peg holes.

That models your question in two ways.
In the first order does not matter. In the second order matters.

9. ## Re: Set of points, 3 objects, get all possible combinations

Hi Plato,

Thanks for that reply. I understand now.

That's alot of combinations! I need this for a input to a program that will take 30 hours (approx) to run per combination so I'm trying to think of ways to reduce the number of combinations that I need to run without comprising the statistical validity of what I'm doing....A friend of mine suggested a fractional factorial analysis....Do you think this is applicable and could you possible point me in the direction of where I would begin such a design?

Thanks again.

10. ## Re: Set of points, 3 objects, get all possible combinations

kerrymaid,
Choosing 2 objects from 12 and choosing 10 objects from 12 are effectively the same thing, so they should give the same number of combinations. When you choose 2, you're not choosing 10, so you're just dividing the 12 objects into two groups.