10,000 combinations.
The number of four-digit combinations is 10,000 .Stick a '3' before each of them, and you have all the possible 5-digit combinations that start with 3.There are 10,000 of them. They run from 30,000 to 39,999 .
The question is incorrect since there are 210 - 1 possible combinations. The digit 0 can be in the combination or out. That gives 2 ways. With each, the digit 1 can be in the combination or out - 2*2 = 22 ways. With each, the digit 2 can be in the combination or out = 23 ways. With each, the digit 3 can be in the combination or out = 24 ways. etc. So 210 ways in all except that one of them is the null combination. Now 210 = 1024 so there are only 1023 combinations. If, instead, you allow the digits to be used many times, there is no limit to the number of combinations.
Each digit can appear in each of the 4 positions. There are 9 digits, therefore there are 9⁴ = 6561 such combinations.
There are ten combinations: one each where one of the ten digits, 0-9, is excluded.
Only 1. In a combination, the order of the numbers does not matter. So 102478 is the same combination as 104782 or 407812 etc.
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
factorial six tat is 6*5*4*3*2*1=720 combinations
It depends. If you can only use each number once, there are 720 combinations. If you can use numbers multiple times, then there are 1000 combinations, by using all numbers from 000 to 999.
The number of four-digit combinations is 10,000 .Stick a '3' before each of them, and you have all the possible 5-digit combinations that start with 3.There are 10,000 of them. They run from 30,000 to 39,999 .
The question is incorrect since there are 210 - 1 possible combinations. The digit 0 can be in the combination or out. That gives 2 ways. With each, the digit 1 can be in the combination or out - 2*2 = 22 ways. With each, the digit 2 can be in the combination or out = 23 ways. With each, the digit 3 can be in the combination or out = 24 ways. etc. So 210 ways in all except that one of them is the null combination. Now 210 = 1024 so there are only 1023 combinations. If, instead, you allow the digits to be used many times, there is no limit to the number of combinations.
Each digit can appear in each of the 4 positions. There are 9 digits, therefore there are 9⁴ = 6561 such combinations.
about 12
the place of each digit help the value of the number by using your multuplication
The answer will depend on how many digits there are in each of the 30 numbers. If the 30 numbers are all 6-digit numbers then the answer is NONE! If the 30 numbers are the first 30 counting numbers then there are 126 combinations of five 1-digit numbers, 1764 combinations of three 1-digit numbers and one 2-digit number, and 1710 combinations of one 1-digit number and two 2-digit numbers. That makes a total of 3600 5-digit combinations.
There are ten combinations: one each where one of the ten digits, 0-9, is excluded.
There are two choices (number 2 or number 3) for each of the five digits. The number of combinations is therefore, 2 x 2 x 2 x 2 x 2 = 32
The short answer is 1000. This is very easy to visualise: Simply consider each number in the combination to be a digit in a decimal number. We then end up with a three-digit number. Such a three-digit number ranges in value from 000 to 999, or 1000 unique combinations.