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kexman
04-09-2010, 03:08 PM
I feel like this is a simple problem, but I can not figure it out. It is not even important, but it will bug me all weekend. I want to know how many distinct combinations I can get from the following scenario:

Simple Case: 4 colors: Red, green, yellow, cyan.

This I can do...4! (4x3x2x1= 24)

My question is how many combinations if I have 8 sets of these colors

(R, G, Y, C);(R, G, Y, C);(R, G, Y, C);(R, G, Y, C);(R, G, Y, C);(R, G, Y, C);(R, G, Y, C);(R, G, Y, C)

and each is independent:

It is not 32 factorial since 6 greens will look the same regardless of which 6 greens are used.

What is the simple equation I am missing? Thanks in advance.

For those wondering about the question it stems from the Brainbow Mouse which can label neurons in many different colors using fluorescent colors. The images are outstanding for those interested (http://www.livescience.com/animals/071031-brainbow.html)

In reality they can see about 100 distinct colors...but we were wondering how to solve the math problem above out of curiosity. thanks

InSpades
04-09-2010, 03:48 PM
So you have 8 sets of 4 different objects and you want to know how many different orders you could put them in?

Reasonably sure that would be 32!/(8!^4) or something along those lines.

The 32! is treating each object as unique. While dividing by 8! 4 times is to take into account that in actuallity there is no difference between some of the objects.

InSpades
04-09-2010, 03:52 PM
You can think of it on a smaller scale... imagine you have 3 Blues, 1 Green and 1 Red. How many ways can you arrange them? If they were all unique you'd say 5! = 120. There are 3! different ways to arrange just the Blues (ignoring the Red and Green placements). All of these ways are actually identical (since we can't tell a difference between the Blues) so only 1 out of every 6 is actually unique. So we have 5!/3! = 20 unique orders. Which makes sense because if we just arranged the Green and Red in 2 of the 5 possible places we have 5 spots for the Green and then 4 remaining spots for the Red (5x4 = 20).

For the record I get about 9.96*10^16 or so.

hurleyfor3
04-09-2010, 04:32 PM
I agree with InSpades.

Pretend you have eight tiles of each color numbered 1 through 8. If you take one arrangement of the whole set, and look at single colors individually, that color's tiles can be arranged in 8! possible ways. This is true for all four colors. So the total number of combinations is 32! divided by (8! to the fourth power).

darthur
04-11-2010, 12:30 AM
This is a pretty famous problem - enough so that the answer (which was said earlier) appears on wikipedia:

http://en.wikipedia.org/wiki/Binomial_coefficient#Generalization_to_multinomial s

It's a little hard to read though.