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I have a maths issue which I think can be solved with factorials. But it's a long time since I used them for anything practical and Internet searches just seem to cover simple scenarios.
Anyway, I have all the letters of the alphabet A-Z (just in upper case) and all 10 digits 0-9. So a total of 36 characters in total.
How many 4 characters combinations exist using these 36 characters
e.g. Starting with
AAAA
AAAB
.
.
AAA0
AAA1
.
etc
Any ideas? Is it possible with factorials or am I using the wrong sort of Maths?
Thanks
Not sure it's factorials as you're not eliminating an option for column b,c,d etc.
I think it's (4^36) * 4 = 18889465931478580854784 but I'm not 100% sure.
Edit: The above is rubbish.
It's (36^4) - So 1679616 is probably more correct.
Here, the internet can do it for you!
http://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html
1679616 is correct.
36^4 = 1679616
Simples.
A Mk5 5 Automatic Sequence Computer would have figured this out for you in no time.
Admittedly the output circuits may have needed modified.
Solution definitely uses factorials 36 choose 4 - I cannot recall the "formula" off the top of my head but in my defense it is 35 years ago - it does have exclamation marks in it though 🙂
Note I am assuming AAAB is counted as different than AABA for example
36^4 is correct.
Compare it with justbhaving the numbers 0-9. We know there are 10,000 possible combinations, including 0000. This is of course 10^4. Just to be sure, if it were binary, there would be 2^4 combinations.
Just re-read this and you would use factorials if you could choose each number / letter only once each time. However you can use the same character more than once so it's the 36 ^ 4 answer as given some time ago by others.
True formthe factorials if you can use each character only once. Imaginge you have only one character, this could have 36 values. Then if there are 2 characters, then for every character there are 35 possibilities for the second character for everyone of those 36 first characters. Sow we can then repeat each ot these with 34 possible third characters, hence 36 x 35 x 34 etc...
You can work out pretty much any combination of characters here:
[url= http://jaburrow.co.uk/Julian/CodingExamples ]My site[/url]
Just go to the Passwords section