This is really a question about permuations with repeated elements.
The formula for that is n!/a! b!... where a, b,.. are repeated
So we have 6 elelments with each repeated 2 times. so the number is.
6!/2!2!2!
This is 6x5x4x3x2x1/ 23
so we can cancel out the 4 and the 2.
6x5x3=90
So 90 different numbers.
There are 26 different letters that can be chosen for each letter. There are 10 different numbers that can be chosen for each number. Since each of the numbers/digits that can be chosen for each of the six "spots" are independent events, we can multiply these combinations using the multiplicative rule of probability.combinations = (# of different digits) * (# of different digits) * (# of different digits) * (# of different letters) * (# of different letters) * (# of different letters) = 10 * 10 * 10 * 26 * 26 * 26 = 103 * 263 = 1000 * 17576 = 17,576,000 different combinations.
Possible 5 digit combinations using 5 digits only 1 time is 5! or 5*4*3*2*1 or 120. Using 5 digits where numbers can be used 5 times is 55 or 3125.
To find all the combinations of the numbers 1204 without repeating any number, we can use the formula for permutations without repetition. There are 4 digits in the number 1204, so there are 4 factorial (4!) or 24 possible combinations. These combinations would include 1204, 1240, 1024, 1042, 1402, 1420, 2104, 2140, 2014, 2041, 2401, 2410, 4012, 4021, 4102, 4120, 4201, 4210, 1024, 1042, 1204, 1240, 1402, 1420.
With repeating digits, there are 33 = 27 possible combinations.Without repeating any digits, there are 6 combinations:357375537573735753
120 There are 6 digits in total. The numbers with 3 digits, with all digits distinct from each other, are the permutations of the 6 digits taken 3 at a time, and therefore there are 6*5*4 = 120 of them.
There are 26 different letters that can be chosen for each letter. There are 10 different numbers that can be chosen for each number. Since each of the numbers/digits that can be chosen for each of the six "spots" are independent events, we can multiply these combinations using the multiplicative rule of probability.combinations = (# of different digits) * (# of different digits) * (# of different digits) * (# of different letters) * (# of different letters) * (# of different letters) = 10 * 10 * 10 * 26 * 26 * 26 = 103 * 263 = 1000 * 17576 = 17,576,000 different combinations.
0000 0001 0002 ... 9998 9999
32C3 = 4960
If you allow digits to be repeated (for example, 222 or 992), then there are 9 x 9 x 9 = 729 combinations. If you do not allow digits to be repeated, then there are 9 x 8 x 7 = 504 combinations.
There are 10C4 = 10*9*8*7/(4*3*2*1) combinations of 4 digits that are all different. That is, 210 of them, and I have no inclination to list them.
The order of the digits in a combination does not matter. So 123 is the same as 132 or 312 etc. There are 10 combinations using just one of the digits (3 times). There are 90 combinations using 2 digits (1 once and 1 twice). There are 120 combinations using three different digit. 220 in all.
9
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
89,999 different numbers i guess
If you exclude numbers starting with zero then the first digit must be between 1 and 9 (i.e. 9 combinations). The remaining 9 digits can be any value between 0 and 9 (i.e. 10 combinations).So you can have 9x109 = 9,000,000,000 combinations.
1
Possible 5 digit combinations using 5 digits only 1 time is 5! or 5*4*3*2*1 or 120. Using 5 digits where numbers can be used 5 times is 55 or 3125.