Counting is hard
I'm posting a problem I have the solution to, and understand why that solution makes sense. What I don't understand is why an alternate method doesn't arrive at the same correct answer.
Problem: You have 10 marbles: 4 red. 3 green, 2 white, 1 purple. If you grab 5 marbles, how many combinations have at least 1 white marble?
The solution: There's 10C5 = 252 combos of 5, and 8C5 = 56 combos with no white marbles. So taking the total combos minus the 0-white combos = 196. Makes sense, but why does the following not also work:
Scenario A with 2 whites pre-chosen: 8C3 = 56. Scenario B with 1 white pre-chosen: 8C4 = 70. (I discard the second white here to avoid double-counting with Scenario A, so we still have 8 marbles we are choosing from, and need to choose 4 to combine with the 1 white).
But 56 + 70 doesn't equal 196 so I am sad, what is the flaw in the second method?
2 Replies
You have to multiply the 8C4 by 2C1. You're not double-counting anything by doing that.
Double-counting would be if you said 5(2/10), in which case you'd need to subtract the overlap (aka inclusion-exclusion):
5(2/10)–(8C3)/(10C5) = 7/9
Btw (8C3)/(10C5) is the same as (5C2)/(10C2), which can be thought of as, the 2 white marbles need to be within the 5 selected ones, out of a possible 10C2 places. Or it can be thought of as, with only 2 selections the chance would be 1/(10C2) but with 5 selections there are 5C2 times as many mutually exclusive chances for it to happen.
Your second scenario only counts half of the combos with one white marble. There are indeed 70 combos of four non-white marbles. There are two choices of white marble; imagine putting a 1 on one and a 2 on the other so you can distinguish them. For each of the 70 combos we can choose either W1 or W2, so there are 140 total combos
