# Math Help - permutations with infinite objects

1. ## permutations with infinite objects

Say I take six colors from an infinite pool of red, green and blue colors and I want to figure out how many orders there are for a certain amount of the colors. e.g. with 3 red, 2 green and 1 blue you might pick them in these orders: RRRGB, RRGRB, etc. and I want to know the total number of orders of those colors that exist. How do I do this? I have very little knowledge of math so please try to make it understandable, or just give me a formula if that's too difficult.

2. ## Re: permutations with infinite objects

Originally Posted by khorven
Say I take six colors from an infinite pool of red, green and blue colors and I want to figure out how many orders there are for a certain amount of the colors. e.g. with 3 red, 2 green and 1 blue you might pick them in these orders: RRRGB, RRGRB, etc. and I want to know the total number of orders of those colors that exist. How do I do this? I have very little knowledge of math so please try to make it understandable, or just give me a formula if that's too difficult.
If you say that $RRRGGB~\&~RRGBGR$ are different orders even though the content is the same then the answer is $3^6$.
Now that includes the possibility of orders of just one color.

3. ## Re: permutations with infinite objects

You are looking for all of the possible permutations of the letters $R,R,R,G,G,B$. So, you have six positions for letters. Out of those six, you will put an R in three of them. So, you have 6 choices for the first position, 5 choices for the second, and 4 choices for the third. So, that is 6*5*4 different possibilities for choosing the positions of the three R's in order. Example: we choose positions 4, 1, 2. However, we could just as easily chosen 1, 2, 4 and we would have red in the same three positions. So, how many different ways could we choose the three positions for red? Out of the numbers 1,2,4, we choose one of them to be the first number (3 choices), two numbers are left for the second number (2 choices) and one number is left for the third number (1 choice). So, there are 3*2*1 different orders of the three numbers. So, we counted the number of ways to choose the positions that are red 3*2*1 times. So, the number of different possibilities to choose three R's in any order is (6*5*4)/(3*2*1). Next, we have three positions left. We want to choose one of them to be blue. So, we have three choices. Then, we have two positions left, and we want both of them to be green, so we have only one choice for those two positions.. The total number of orders for R, R, R, G, G, B is $\dfrac{6\cdot 5 \cdot 4}{3 \cdot 2 \cdot 1} \cdot 3 = 60$.

4. ## Re: permutations with infinite objects

If you want a general formula, suppose you choose $n$ colors with $r_1$ of them red, $r_2$ of them green, and $r_3$ of them blue so $r_1+r_2+r_3 = n$. Then the total number of orders is $\binom{n}{r_1,r_2,r_3} = \dfrac{n!}{(r_1)!(r_2)!(r_3)!}$. So, in the example you had it would be $\dfrac{6!}{3!2!1!} = \dfrac{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{(3\cdot 2\cdot 1)\cdot (2\cdot 1)\cdot (1)} = 60$

5. ## Re: permutations with infinite objects

Thank you!

Originally Posted by Plato
If you say that $RRRGGB~\&~RRGBGR$ are different orders even though the content is the same then the answer is $3^6$.
Now that includes the possibility of orders of just one color.
I phrased the question badly. I know this is for all the possible orders for any color in any position, but I wanted to know what SlipEternal posted. Thanks though.

6. ## Re: permutations with infinite objects

Hello, khorven!

I hope you understand "factorial" notation.

Suppose we want $RRRRGGGBB$ in all possible permutations.
(4 Reds, 3 Greens, 2 Blues.)

If they asked for nine different colors,
. . (one each of: Red, Orange, Yellow, Green, Blue, Purple, Black, White, Brown)
there would be $9!$ possible permutations.

But there are 4 identical Reds.
They can be swapped around in $4!$ ways
. . and produce the same permutation.
So we must divide by $4!.$

And there are 3 identical Greens.
They can be swapped around in $3!$ ways
. . and produce the same permutation.
So we must divide by $3!.$

And there are 2 identical Blues.
They can be swapped around in $2!$ ways
. . and produce the same permutation.
So we must divide by $2!.$

Therefore, there are: . $\frac{9!}{4!\,3!\,2!} \:=\:1260$ permutations.

7. ## Re: permutations with infinite objects

That helps me understand, thank you! I can't totally follow you because of my math knowledge (like I don't even know why we multiply the possibilities even though I know it's right) but I think this is all I need.

8. ## Re: permutations with infinite objects

Originally Posted by khorven
That helps me understand, thank you! I can't totally follow you because of my math knowledge (like I don't even know why we multiply the possibilities even though I know it's right) but I think this is all I need.
Let's consider a single choice of positions for red:

R,R,R,_,_,_

For this choice for the positions of the reds, we have three spots left to choose where to put the 2 green and 1 blue. They can be in any of the following orders: G,G,B or G,B,G or B,G,G. There are three possible orders for those three colors.

What about for a different choice? Let's say we look at R,R,_,R,_,_. We still have three spots left to choose where to put the 2 green and 1 blue. And again, they are the same three orders (but now interspersed with a single red). In fact, no matter where we choose to put the three reds, we are left with three spots to put 2 green and 1 blue. So, let's add up all of the different orders.

We get three orders from R,R,R,_,_,_
We get three orders from R,R,_,R,_,_
We get three orders from ...
We get three orders from _,_,_,R,R,R
Adding that up, we get 3*(# of ways to order the three R's in the first place). That's why we multiply.

9. ## Re: permutations with infinite objects

I see. Thanks again!