This is what I was thinking about - what was discussed in my previous posts
"More generally, the formula for finding the number of combinations of k objects you can choose from a set of n objects is:
n!
n_C_k = ----------"
k!(n - k)!
This might clarify things: Daylate, you said “I’m trying to calculate the number of 5 digit numbers that can be formed out of a series of numbers that is 55556 long”. Consider, for analogy, the question of how many 2 digit numbers can be formed out of a series of numbers that is 11 long; call the answer to this question X. Can you not only tell us what X is, but also list out specifically what the precisely X many 2 digit numbers are, so we then understand more generally what sort of thing you are counting?
According to my calculations, the number of two letter words made from a list of 11 letters is 55. The calculations are shown below as Case 2. I did a Case 1 with the number of two letter words that can be made from a list of 3 letters and this came ot to 3, which I think is the correct answer.
Case 1 Case 2
n = 3 11
k = 2 2
n ! = 6 39,916,800
k! = 2 2
(n-k) = 1 9
(n-k)! = 1 362,880
n!/k!*(n-k)! = 3 55
I apologize for the garbled up formatting in my previous post. It was OK when it left here by the SD made its own changes before it got into the posts.
It looks like the Wolfram Alpha procedure works OK. That’s a herkin big number of combinations
BTW, I will be out of town beyond the reach of the Internet until Monday PM.
Evidently the SD hates me - another garbled up format. Sorry.
Anyway, the Wolfram-alpha method seems to work. That’s a herkin big number, something like 10 to the 21st power. About the number of grains of sand on Earth.
Thanks to all who tried to help. This board has some pretty smart folks on tap!
I don’t want calculations. I want you to list all the possibilities explicitly. Give me a list of every possibility, which I can then count up myself.
If you think there are too many to easily list, feel free to replace “11” with, say, “4”.
(I originally picked “11” because you had previously mentioned “5 digit numbers that can be formed out of a series of numbers that is 55556 long” and I wished to test how “digit” was being interpreted. But now you have switched to discussing “letters” and “words” instead of “digits” and “numbers”…)
The reason I am pushing on this is that I fear you may be blindly applying a formula without understanding whether it is the appropriate formula for what you are actually interested in. [Put another way, do you recognize how the “n!/k!*(n-k)!” formula is derived? Are you sure that derivation is applicable to what you are ultimately interested in counting?].
For example, you said above that you think there are 3 two letter words which can be made from a list of 2 letters. Many of us would normally say there are 9 two letter words which can be made from a list of 2 letters: AA, AB, AC, BA, BB, BC, CA, CB, and CC. You’ve indicated that you don’t consider order important, so perhaps we should say only that there are 6 two letter words which can be made from a list of 2 letters: AA, AB, AC, BB, BC, and CC. If we furthermore tossed out AA, BB, and CC on the grounds that they have one letter duplicated twice, we would get down to only 3 items, but you indicated previously that you wished to allow (at least, up to two) duplicates. It is unclear how to reconcile this difference.
These are the kinds of confusions (for us in understanding your intent) which could potentially be cleared up if you would explicitly list the items you are counting in some example.
Small (but repeated) typo corrected in bold.
https://en.wikipedia.org/wiki/Combination
This link (good old Wikipedia) contains the formula that I am using. n!/(k!*(n-k)!).
That seemed to me to be what I was after.
BTW, I will be out of town until Monday PM so won’t be able access this Board until then.
You are not answering my question (“I don’t want calculations. I want you to list all the possibilities explicitly. Give me a list of every possibility, which I can then count up myself.”). You are answering some other question.
Yes, we know that that’s where the formula comes from. We still don’t know what it is you’re trying to do, to tell if it matches.