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genjix China. Nov 04 2009 00:12. Posts 2677 | | |
Hey,
I often forget this stuff and have to look it up. So am making a note here for myself.
Combinations are unordered giving the number of combinations for a prescribed size.
C(n,r) = n! / r! (n - r)!
An example would be a 2 card Poker hand from a pack of 52 cards:
C(52,2) = 52! / 2! 50!
Taking 2 letters from ABCD, we get 6. This is because AB = BA... Combinations are unordered.
Permutations are ordered remappings of a set.
P(n,r) = n! / (n - r)!
= r! C(n,r)
Given a 3 card flop, total orderings will be 3!/(3-3)! = 3!. This makes sense cos we have 3 positions for the first card (1st, 2nd, 3rd), then only 2 for the next card and 1 for the last... n(n-1)(n-2)...2x1x0
ABC
ACB
BAC
CAB
BCA
CBA
note:
With replacement, the combination formula becomes:
C(n,k) = (n + k - 1) / k! (n - 1)!
Multiple elements can be represented using 0's and 1's like:
11 0 111 0 111 0 1
A | B | C | D
where the 0's are separators for the slots (ABCD). Therefore the combinations will just be the total permutations of all 12 0's and 1's together. Or C(4,12). 4 choices but 12 samples.
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| If you wish to make an apple pie from scratch, you must first invent the universe. | Last edit: 05/11/2009 17:41 |
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morph1 Sierra Leone. Nov 05 2009 07:37. Posts 2352 | | |
| | On November 03 2009 23:12 genjix wrote:
Hey,
I often forget this stuff and have to look it up. So am making a note here for myself.
....
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it's his blog.. he can post whatever he wants |
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