I am assuming you mean 3-number combinations rather than 3 digit combinations. Otherwise you have to treat 21 as a 2-digit number and equate it to 1-and-2. There are 21C3 combinations = 21*20*19/(3*2*1) = 7980 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 .
6 of them.
There are 15180 combinations.
There are 12C4 4 NUMBER combinations. And that equals 12*11*10*9/(4/3/2/1) = 495 combinations. However, some of these, although 4 number combinations consist of 7 digits eg 1, 10, 11, and 12. Are you really sure you want 4-DIGIT combinations?
I am assuming you mean 3-number combinations rather than 3 digit combinations. Otherwise you have to treat 21 as a 2-digit number and equate it to 1-and-2. There are 21C3 combinations = 21*20*19/(3*2*1) = 7980 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 .
6 of them.
the answer is = first 2-digit number by using 48= 28,82 and in 3 digit is=282,228,822,822
The number of possible 3 digit combinations you can make out of 1-9 with outrepeated digits is:9C3 = 9!/(3!(9-3)!) = 84
There are 15180 combinations.
For the first digit you have 5 options, whichever you choose for the first digit, you have 4 options for the second digit, etc.; so the number of combinations is 5 x 4 x 3 x 2.For the first digit you have 5 options, whichever you choose for the first digit, you have 4 options for the second digit, etc.; so the number of combinations is 5 x 4 x 3 x 2.For the first digit you have 5 options, whichever you choose for the first digit, you have 4 options for the second digit, etc.; so the number of combinations is 5 x 4 x 3 x 2.For the first digit you have 5 options, whichever you choose for the first digit, you have 4 options for the second digit, etc.; so the number of combinations is 5 x 4 x 3 x 2.
How many four digit combinations can be made from the number nine? Example, 1+1+2+5=9.
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.
To calculate the number of 3-digit combinations that can be made from the numbers 1-9, we can use the formula for permutations. Since repetition is allowed, we use the formula for permutations with repetition, which is n^r, where n is the total number of options (10 in this case) and r is the number of digits in each combination (3 in this case). Therefore, the total number of 3-digit combinations that can be made from the numbers 1-9 is 10^3 = 1000.
A 14-digit number is typically referred to as a "14-digit number." In mathematics, numbers are generally identified by the number of digits they contain, such as a 3-digit number, 7-digit number, etc. In computing, a 14-digit number may also be referred to as a "14-digit integer" or "14-digit numerical value."
There are 12C4 4 NUMBER combinations. And that equals 12*11*10*9/(4/3/2/1) = 495 combinations. However, some of these, although 4 number combinations consist of 7 digits eg 1, 10, 11, and 12. Are you really sure you want 4-DIGIT combinations?