Assuming the digits are not repeated, there are four combinations:123, 124, 134 and 234.
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.
Oh, dude, you're making me do math now? Alright, so if you have three numbers and you're asking how many combinations you can make with those three numbers, it's like a little math puzzle. Each number can be used multiple times, so it's like a little party for those numbers. The total number of combinations you can make with three numbers is 27. That's like having 27 different outfits to choose from for a night out, but with numbers.
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 .
the answer is = first 2-digit number by using 48= 28,82 and in 3 digit is=282,228,822,822
6 of them.
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.
To calculate the number of 4-digit combinations using the digits 1, 3, 5, and 7 exactly once each, we can use the permutation formula. There are 4 choices for the first digit, 3 choices for the second digit, 2 choices for the third digit, and 1 choice for the fourth digit. Therefore, the total number of combinations is 4 x 3 x 2 x 1 = 24. So, there are 24 possible 4-digit combinations using the digits 1, 3, 5, and 7 exactly once each.
Oh, dude, you're making me do math now? Alright, so if you have three numbers and you're asking how many combinations you can make with those three numbers, it's like a little math puzzle. Each number can be used multiple times, so it's like a little party for those numbers. The total number of combinations you can make with three numbers is 27. That's like having 27 different outfits to choose from for a night out, but with numbers.
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.