How many different 3-digit combinations can I make using the numbers 0-9? and what are they?
2006-10-19 19:16:06 · 7 answers · asked by E.F. Landeros 3 in Science & Mathematics ➔ Mathematics
I think 1000
000-999
2006-10-19 19:22:24 · answer #1 · answered by RichUnclePennybags 4 · 0⤊ 0⤋
So you got 10 numbers 0 to 9. So its 10*9*8=720 if you use numbers only once in that 3 digit number. I think.
2006-10-19 19:27:28 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Assuming you can only use each digit once in each 3-letter combination...
012
013
014
015
016
017
018
019
021
023
024
025
026
027
028
029
031
032
034
035
036
037
038
039
041
042
043
045
046
047
048
049
051
052
053
054
056
057
058
059
061
062
063
064
065
067
068
069
071
072
073
074
075
076
078
079
081
082
083
084
085
086
087
089
091
092
093
094
095
096
097
098
So... if you follow this pattern... There are 72 possible ending for each beginning number. The first number can be any of the ten digits.
Therefore, there are 720 possible combinations.
Hope this helps. :-)
2006-10-19 19:31:38 · answer #3 · answered by Anonymous · 0⤊ 0⤋
The #'s 000 through 999 . . . that would make 1000 combinations counting 000. they would be 000, 001, 002, 003, 004 . . . 101, 102, 103 . . .999. I hope you get the idea, cause I am not going to list them all.
2006-10-19 19:23:52 · answer #4 · answered by manderso750 2 · 0⤊ 0⤋
9!/3! X 6! i think
2006-10-19 21:54:13 · answer #5 · answered by Anonymous · 0⤊ 0⤋
This question of yours wasn`t so clear. It didn`t mention whether repitition is allowed or not.
But the answer is
1) when repitition is allowed - 900.
2) repitition isn`t - 728
2006-10-19 20:36:02 · answer #6 · answered by UtAkArSh 2 · 0⤊ 0⤋
there is a simple formula for this. however many #'s u r working w/ times it by itself, so, 10 times 10
2006-10-19 20:49:05 · answer #7 · answered by Anonymous · 0⤊ 0⤋