Blue Proof
How many ways are there to arrange m red balls and n blue balls in a row?
We do not know which is bigger, m or n. The red balls are not distinct, ditto for the blue balls. Meaning that if I swap around two of the red balls, it still counts as the same.
I think there is some sort of formula for this, but I have no idea what it is called and how to look for it. I need this method to solve a larger question in a college problem set. It would be awesome if there is some sort of proof, too, though I don’t need this for the problem set.
The number of different combinations of k items, chosen from p possibilities, is denoted
pCk or C(p, k) and can be found by the formula
pCk = p! / [(p-k)! k!]
In this case, the row contains m+n balls, and we want to know how many ways we can pick the n positions for the blue balls. The unpicked ones will be the red balls. So the number of different arrangements is
(m+n)Cn = (m+n)! / (m! n!)
Notice that if, instead, we picked the m positions for the red balls and let the unpicked positions be the blue ones, we’d get the same formula. Looking at the problem a different way doesn’t change the answer.
If you’re familiar with the difference between combinations (where order doesn’t matter) and permutations (where order does matter, referring to how we “permute” or re-order the choices), this may seem like an odd formula to use, because we are considering the different sequences of the same red and blue balls significant. But the key here is to regard ourselves as choosing a set of positions in the row to be one color. The order in which we choose the positions doesn’t matter; only the set of positions chosen does. That’s why the combinations formula, not the permutations one, applies to this case.
blue proof 2