# Probability equation question

1. Sep 23, 2011

### zeion

1. The problem statement, all variables and given/known data

I don't understand the equation for drawing with replacement and considering results to be unordered.

So with replacement and no attention to order all we see is the number of each color drawn. So this is represented by o's and x's where o's is the number of that color drawn and x is a separator for the next color.

If # of different colors is n and # of draws is k then # of o's is k and # of x's is n - 1.
So now I need to find all the possible permutation of this set of x's and o's.

I don't understand why it is k + n - 1 choose n - 1 (?)

(k + n - 1)! / (n - 1)! (k + n - 1 - (n - 1))! (?)

2. Relevant equations

3. The attempt at a solution

2. Sep 23, 2011

### CompuChip

I like to think of "a choose b" as "the number of ways to pick b places from a total of a".
In that case, it is easy to see if you imagine the following; suppose you have k o's and (n - 1) x's. Imagine k + n - 1 empty spots on the table, which is just enough to hold all the o's and x's. Now you can fix any order, if you pick (n - 1) places where all the x's will go. Since all the x's are indistinguishable, and the all o's are too, if you have n - 1 places for the x's the o's will have to in the remaining k places. So you want to answer the question "in how many ways can I pick n - 1 spots from the total of k + n - 1 to place my x's on?" which is -- almost by definition -- (k + n - 1) choose (n - 1)

3. Sep 24, 2011

### zeion

So if I do something like 2 choose 3 I get a negative.
Does that mean it's impossible?

4. Sep 25, 2011

### CompuChip

If you do 2 choose 3 you should get zero: How many ways are there to pick 3 objects from a set of 2?
(That's also what my "calculator" gives, how do you get a negative?)

5. Sep 25, 2011

### uart

I think he's referring to getting the factorial of a negative number on the denominator. This is consistent with zero combinations anyway.

6. Sep 25, 2011

### HallsofIvy

The formula for "n choose i" is $_nC_i= \begin{pmatrix}n \\ i\end{pmatrix}= \frac{n!}{i!(n- i)!}$.

For "2 choose 3" it is $\frac{2!}{3!(-1)!}$ but (-1)! is undefined. It is a convention that assigns a value of 0 to it.

But the answer to your question is "yes". It should be obvious that if you only have two objects, you cannot choose three of them!