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In case anyone is curious as to how to do some elementary statistics, in this case combinatorics, here is a quick little write up I did to explain the "n choose k" paradigm.
Here is the is the question (from my cousin):
> Math people
Can you help me ??
Let's say
Hypothetically
You have 7 numbers
Between 1-99
The first 6 cant repeat but the 7th can
How many outcomes are possible?
This isnt a riddle or anything I just dont know how to do the math on something I'm trying to figure out
Here is my answer:
Might benefit from a discussion of why `nCk = n!/(k!(n-k)!)`. I don't use combinatorics often enough to have it memorised, so I re-derive the `nCk` and `nPk` formulae every time I need them. Here's my reasoning:
- If you want to arrange `n` objects, you can choose any of `n` for the first, any of the `n-1` remaining for the second, ... down to one for the `n`th. So n objects can be arranged in `n! = n(n-1)...1` ways.
- If you want to choose `k` of `n` objects, you can do this by ordering the `n` objects and taking the first `k`. So initially `nCk = n!`
- However, you don't care about ordering of the chosen objects. If you choose two of `{1, 2, 3, 4}`, the orderings `{1, 2, 3, 4}` and `{2, 1, 3, 4}` should not be treated as distinct. Since there are `k!` ways to order the `k` objects you chose, you need to divide that out. Now we have `nCk = n!/k!`
- Equally, you don't care about ordering of the non-chosen objects. In the above example, `{1, 2, 3, 4}` and `{1, 2, 4, 3}` should not be treated as distinct. There are `n-k` of these objects, so `(n-k)!` ways to order them, which again should be divided out. Finally, we arrive at `nCk = n!/(k!(n-k)!)`
