To calculate the total number of possible 4-digit codes using the numbers 1, 2, and 3, we can use the fundamental principle of counting. Since we have 3 choices for each digit and we are choosing 4 digits, the total number of possible codes is 3 x 3 x 3 x 3 = 81. Therefore, there are 81 possible 4-digit codes that can be formed using the numbers 1, 2, and 3.
Since you've only given us 3 digits to put in 4 places, we have to assume
that it's OK to use the same digit more than once in a code. So we're also
going to assume that there's no limit on repetition.
The first digit can be any one of 3. For each one . . .
The second digit can be any one of 3. For each one . . .
The third digit can be any one of 3. For each one . . .
The fourth digit can be any one of 3.
Total possibilities = 3 x 3 x 3 x 3 = 81 possible codes.
Which are:
1111, 1112, 1113, 1121, 1122, 1123, 1131, 1132, 1133, 1211, 1212, 1213, 1221, 1222, 1223, 1231, 1232, 1233, 1311, 1312, 1313, 1321, 1322, 1323, 1331, 1332, 1333, 2111, 2112, 2113, 2121, 2122, 2123, 2131, 2132, 2133, 2211, 2212, 2213, 2221, 2222, 2223, 2231, 2232, 2233, 2311, 2312, 2313, 2321, 2322, 2323, 2331, 2332, 2333, 3111, 3112, 3113, 3121, 3122, 3123, 3131, 3132, 3133, 3211, 3212, 3213, 3221, 3222, 3223, 3231, 3232, 3233, 3311, 3312, 3313, 3321, 3322, 3323, 3331, 3332, 3333
1111, 1112, 1113, 1121, 1122, 1123, 1131, 1132, 1133, 1211, 1212, 1213, 1221, 1222, 1223, 1231, 1232, 1233, 1311, 1312, 1313, 1321, 1322, 1323, 1331, 1332, 1333, 2111, 2112, 2113, 2121, 2122, 2123, 2131, 2132, 2133, 2211, 2212, 2213, 2221, 2222, 2223, 2231, 2232, 2233, 2311, 2312, 2313, 2321, 2322, 2323, 2331, 2332, 2333, 3111, 3112, 3113, 3121, 3122, 3123, 3131, 3132, 3133, 3211, 3212, 3213, 3221, 3222, 3223, 3231, 3232, 3233, 3311, 3312, 3313, 3321, 3322, 3323, 3331, 3332, 3333
There are 210 4 digit combinations and 5040 different 4 digit codes.
Any 4 from 10 in any order = 10 x 9 x 8 x 7 = 5040
There are 4 possible numbers if the digits are not repeated; 18 if they are. Those are 3-digit numbers, assuming that zero would not be a leading digit. If zero is allowed for a leading digit, then you can have 6 for the non repeated, and 27 if repetition.
10,000
There are 256. Some of them are: 1111, 1112, 1113, 1114, 1121, ... 2111, 2112, 2113, ... 3111, 3112, ... 4111, etc.
There are 210 4 digit combinations and 5040 different 4 digit codes.
There are 10000 such codes. Each of the numbers 0-9 can be in the first position. With each such first digit, each of the numbers 0-9 can be in the second position. With each such pair of the first two digits, each of the numbers 0-9 can be in the third position. etc.
Any 4 from 10 in any order = 10 x 9 x 8 x 7 = 5040
1,000 of them. The list of possibilities will look exactly like the counting numbers from 000 to 999 .
1000
There are 4 possible numbers if the digits are not repeated; 18 if they are. Those are 3-digit numbers, assuming that zero would not be a leading digit. If zero is allowed for a leading digit, then you can have 6 for the non repeated, and 27 if repetition.
The possible 4 digit codes using the numbers 0-9 are every number between 0 and 9999. For numbers that have less than 4 digits, just precede the number with 0's. 10,000 possibilities
Total number of possible 3-digit numbers = 9!x10!10!
10,000
103 = 1000 in all.
9999999999 100000000
Using the digits of 1345678, there are 210 three digit numbers in which no digit is repeated.