Is there anyone mathmatically literate that has or can fiqure out every possible 4 digit combination for 0-9? I reallly need your help.
2006-09-20 03:32:11 · 12 answers · asked by bena042000 1 in Science & Mathematics ➔ Mathematics
I have looked on the internet on how to figure it out, but I would consider myself very mathmatically challage. Is there some one willing do give me all the 4 digit combinations for 0-9
2006-09-20 03:41:07 · update #1
all #'s can be used more then once.
2006-09-20 03:42:14 · update #2
email me at bena042000@yahoo.com if you are willing
2006-09-20 03:59:15 · update #3
If you have 4 possible digits: A, B, C, and D.
then A has ten possibilities, B has ten possibilities, C has ten possibilities, and D has ten possibilities.
Therefore, 10 *10 * 10 *10 = 10000
2006-09-20 04:01:33 · answer #1 · answered by MB_Bailey 3 · 1⤊ 4⤋
There are a lot of combinations!
If you allow duplicates (eg 1111) then there are 10 possibilities for the first digit, 10 for the second etc, making 10,000 diferent combinations in total.
If you want to see the combinations then write out all the numbers from 0000 to 9999
If, however, you can only use each digit once, then there are 10 possibilities for the first digit, 9 for the second, 8 for the third and 7 for the fourth - a total of 10x9x8x7=5040 different combinations.
2006-09-20 10:41:19 · answer #2 · answered by robcraine 4 · 0⤊ 0⤋
More information is needed. Can any digit be used more than once in a combination?
If not, then the smallest 4 digit combination is 1023 (or possibly 0123) and the highest is 9876.
If so, then can you include 0001 and 9999 as four digit combinations? With leading zeros, you could have 0000 to 9999 making 10,000 combinations.
For non-repeating combinations, you would have 10*9*8*7 = 5040.
What about Non-integers? Is 0.001 allowed?
Negative numbers? Is -1234 okay?
Is 12/34 a four digit combination?
2006-09-20 10:37:14 · answer #3 · answered by Richard 7 · 5⤊ 0⤋
That would be 10000 combinations right? Or do you mean using each digit only once? By the way, 10*9*8*7= 5040.
Ok then, 10000 combinations. Why? Because the numbers 0000 to 9999 cover every posible combiation.
2006-09-20 10:35:26 · answer #4 · answered by pito16places 3 · 1⤊ 0⤋
If you're allowed to repeat numbers 10000. If not take 10 x 9 x 8 x 7.
2006-09-20 10:36:06 · answer #5 · answered by Kris B 5 · 0⤊ 0⤋
If repeating digits are allowed (4442), it's a simple permutation: ten possibilities in four slots...n^r = 10^4 = 10,000 possible results.
If repeating digits aren't allowed, the permutation looks like:
n! / (n-r)!
where n is the number of objects to choose from, and r is the number of objects to choose.
10! / (10 - 4)! = 10! / 6! = 5040 possible results
2006-09-20 10:49:26 · answer #6 · answered by Professor 3 · 0⤊ 0⤋
It takes a really long time to open one of those combination locks by trying all the combinations. There are 10,000 of them starting with 0000 and ending with 9999.
2006-09-20 10:39:57 · answer #7 · answered by Anonymous · 0⤊ 0⤋
I believe 10,000 is correct ...1 for each of the numbers from 0 to 9999
2006-09-20 10:37:58 · answer #8 · answered by The Cheminator 5 · 0⤊ 0⤋
there are 10,000 combinations possible, since 10 to the 4th power is equal to ten-thousand.
2006-09-20 10:44:12 · answer #9 · answered by Em 2 · 0⤊ 0⤋
I dont know but I would suggest that you go to a college or university and ask for a mathematician.
2006-09-20 10:41:28 · answer #10 · answered by diamondswintergreen 2 · 0⤊ 0⤋