8-digit combination

re
we need to agree on the vocabulary and, incidentally, know what exactly bebess wants
- combination > the order does not matter
- arrangement > the order matters
- with repetition > an element can be repeated
- without repetition > an element appears at most once
and to be exhaustive, their number is (unless I’m mistaken)
- Combinations > C(10,8) = 45
- Arrangements > 10!/(10-8)! = 1 814 400
- Combinations with repetition > C(10+8-1,8) = 24 310
- Arrangements with repetition 10^8 = 100 000 000
as for writing them all .....
have a good day

ccm81, look, let me explain it to you clearly,
this is about probabilities,
you need 8 positions, and each position takes its value from the list 0,1,2,3,4,5,6,7,8,9 (10 values)

so you have: 10^8 which equals 100,000,000

example:
0 0 0 0 0 0 1 2
^ ^ ^ ^ ^ ^ ^ ^
10 10 10 10 10 10 10 10

10 being the number of possibilities for each number

hello everyone

there are indeed 45 combinations of 8 digits from 10 (from 0 to 9)
with the old notation
C(10,8) = C(10,2) = 10*9/2 = 45

have a good continuation

re
>malaik5
- as long as we're not speaking the same language, we can't understand each other. I suggest you refer back to your senior year classes or Wikipedia for a vocabulary update.
- for me, the discussion is closed
best wishes

45 combinations without order
10^8 or 100,000,000 with order
--
Life is like a box of chocolates, you never know what you're gonna get.

What’s the point?
You just need to know how to count up to 99,999,999...
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Trust does not exclude control.