How do you calculate how many combinations there will be?

[GrandMasterK]

As a simple example. If I have a 4 balls, red, yellow, blue and green, and there are 4 places in which I could place these balls, lets say left, left middle, right middle and right. How do I figure out how many combinations of arrangements there could be? I thought it was multiplying the number of spots by how many different kinds of balls could be in one spot. However that gives me 16, and when I start doing it in my head I realize it's way more then that. If that's two hard to understand then, I want to see how many of these there are:

 

red, yellow, blue, green

 

yellow, blue, green, red

 

blue, green, red, yellow

 

etc etc

 

After that, how bout a bigger kicker, what if you could have the same color in more then one spot, but limited to those 4 colors?

[the tree]

[math]^{n}C_{k}[/math] gives the amount of ways that n items can be put into k places. It's worked out as [math]^{n}C_{k}=\frac{n!}{(n-k)!k!}[/math]

[RyanJ]

I believe the equation you are looking for is:

 

[math]\frac{n!}{(n - r)!r!}[/math]

 

Where n is the number of objects (In your case 4) and r is the number of objects to be chosen (4 again)

 

Substitute the values and you'll have an answer

 

Cheers,

 

Ryan Jones

[Rasori]

I don't see how that works out. From this example, you'd end up with 4! (or 24) over 4! (because (4-4)! = 0! = 1) * 4!

 

That would be 24 / 24 = 1, no? And there isn't 1 possibility here. I'm probably doing the problem wrong, though, because I don't deal with factorials often yet. Can someone tell me what's up?

[RyanJ]

I don't see how that works out. From this example' date=' you'd end up with 4! (or 24) over 4! (because (4-4)! = 0! = 1) * 4!

 

That would be 24 / 24 = 1, no? And there isn't 1 possibility here. I'm probably doing the problem wrong, though, because I don't deal with factorials often yet. Can someone tell me what's up?[/quote']

 

Maybe it should have been the permitation formula and not the combination formula.

 

[math]

\frac{n!}{(n - r)!}

[/math]

 

http://en.wikipedia.org/wiki/Combinations_and_permutations

 

Cheers,

 

Ryan Jones

[Rasori]

That would make it 24... let's work this out.

1. RYBG, 2. RYGB, 3. RBYG, 4. RBGY, 5. RGYB, 6. RGBY, 7. YRGB, 8. YRBG, 9. YGRB, 10. YGBR, 11. YBGR, 12. YBRG, 13. BRGY, 14. BRYG, 15. BGYR, 16. BGRY, 17. BYGR, 18. BYRG, 19. GRYB, 20. GRBY, 21. GYRB, 22. GYBR, 23. GBRY, 24. GBYR.

 

Yeah, that seems to work.

 

It works out to this (I asked my sister for this answer). If you can't repeat, it's number of possibilities for each slot, working it down. The equation works good, but if you can't remember equations well (she can't), it's number of possibilities (n) * (n-1) * (n-2) and so on, where the first term is for the first slot, and the second (n-1) is the second slot, continuing down the line.

 

If it can be repeated, it's just n^x where ^x is the number of slots.

 

Edit: My first answer for repeatable situations was completely wrong.

[Tartaglia]

nCr is the number of ways of choosing r objects from n objects when the order doesn't matter

nPr is the number of ways of choosing r objects from n when the order matters

If you choose 4 from 4 and the order matters then 4P4 will work but it is not an efficient way of thinking about this problem. The answer is of course just 4! = 24

[GrandMasterK]

What does the ! in the equation mean?

 

Anyway I'm not seeing how that equation works. n and r are the same so my answer on the bottom always subtracts itself to 0, making it 4 over 0 which =0, help me out here.

[Tartaglia]

! means factorial

 

ie 4! = 4*3*2*1

 

This problem has been unnecessarily complicated. The number of ways of arranging n different objects in n boxes is just n!