eaakbari wrote:
Well, my logic was
The number of ways of selecting r objects from n different objects is nCr.
We have 8 options and can select either of the 8 as the first - 8C8
One car is gone and one becomes a forbidden choice so we have 7 options and need to select either of the 6 as the first - 7C6
Two cars are gone and two becomes a forbidden choice so we have 6 options and need to select either of the 4 as the first - 6C4
I reckon that my understanding of Combinatorics logic is fundamentally flawed. Appreciate if you could tell me where and what
Yes, I think you need to go through a PnC book from scratch. You can check out the
Veritas Combinatorics and Probability book at amazon.
"The number of ways of selecting r objects from n different objects is nCr" - This is correct. You need to apply it correctly now.
As for this question,
We have 8 options and can select either of the 8 as the first - 8C8
--- We have 8 options and we need to select 1 out of those 8. So n = 8 and r = 1. This gives us 8C1 ways. (Say you selected BlackA)
One car is gone and one becomes a forbidden choice so we have 7 options and need to select either of the 6 as the first - 7C6
--- Now one car is gone and we cannot select another one. Now we have 6 cars to choose from and we again have to select 1. So n = 6 and r = 1. We get 6C1 ways (Say you selected GreenB)
Two cars are gone and two becomes a forbidden choice so we have 6 options and need to select either of the 4 as the first - 6C4
--- Now 2 cars are gone and 2 are forbidden. You can choose out of only 4 cars and you have to select only 1. Again, n = 4 and r = 1 so we get 4C1 (Say you selected RedA)
You select 3 cars in 8C1 * 6C1 * 4C1 ways but there is a flaw here. You selected BlackA, GreenB and RedA.
In another case, you could have selected RedA first, then BlackA and then GreenB. This combination is the same as the previous one but we are counting it separately. The point is we have arranged the cars as first, second and third which we should not do since it is a group.
As I said before, you might want to start with fundamentals from a book.
I made a similar error in the beginning and looking back at it, it might have been because I might have tried to mix in "probability" into the combinatorics problem.
If you treat it as a probability, wouldn't the above posters approach be correct?