Simple enough to solve. The answer is a power of two. Assuming you have two possible digits, say for example, 3 and 4, then you simply have to multiply it by how many numbers you want to get the total number of combinations. Each number can be 3 or 4 in this case, and you have 5 numbers. That's two to the fifth. Five combinations of any two numbers. 2x2x2x2x2. The answer is 32 combinations.
It depends upon whether the numbers can be used more than once. If the numbers can be used more than once, then there are 1,000 possible combinations; if not, then there are 720 possible combinations. ========== Assuming you are talking about integers you can calculate it this way: you can have any one of 9 digits for the first digit (zero is excluded because that would make it only a 2 digit number) You can have any one of 10 digits for the second and any one of 10 digits for the third digit. That gives you 9x10x10 = 900 different combinations for 3 digit numbers (not 1000). If you can include both negative and positive numbers as different numbers you get twice as many (2x900=18000). If you can count decimal numbers as 3 digit numbers (i.e. not restricted to integers) you can have 900 integers, 900 with the form xx.x, 1000 with the form x.xx (if zero is permitted to be the first digit and count as one of the 3 digits) or 900 (if a leading zero is NOT counted as one of the 3 digits). If a leading zero is NOT counted as one of the 3 digits, you could also have 1000 numbers of the form 0.xxx or just .xxx
what is the greatest 4-digit number you can make with the digit 0,1,2,3, and 4
12 numbers
If you want 4-digit numbers, there are 24 of them.
90
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.
You can create several different numbers from the digits 1, 3, 4, and 6 in 1346. If you consider all possible combinations of these four digits, you can form a total of 24 unique four-digit numbers (4! = 24). Additionally, you can also create various three-digit, two-digit, and one-digit numbers, which would increase the total count significantly. However, counting all permutations of all possible lengths gives a comprehensive view of the total combinations.
The number of possible 3 digit combinations you can make out of 1-9 with outrepeated digits is:9C3 = 9!/(3!(9-3)!) = 84
Not sure what a "didget" is. It is possible to make 18 5-digit numbers.
Oh, dude, you're hitting me with the math questions, huh? Alright, so to find the number of 5-digit combinations from 1 to 20, you just do 20^5, which is like 3,200,000. So, yeah, there are 3,200,000 possible 5-digit combinations from 1 to 20.
Oh, dude, let me break it down for you. So, to find the number of 5-digit combinations from 1 to 60, you just do 60 minus 1 plus 1, which gives you 60. So, there are like 60 different 5-digit number combinations you can make from that range. Easy peasy, lemon squeezy!
There are 5*4*3*2 = 120 such numbers.
4578, 4758, 5478, 5748, 5874, 5784, 7584, 7854, 7458, 7548, 8754, 8574
Simple enough to solve. The answer is a power of two. Assuming you have two possible digits, say for example, 3 and 4, then you simply have to multiply it by how many numbers you want to get the total number of combinations. Each number can be 3 or 4 in this case, and you have 5 numbers. That's two to the fifth. Five combinations of any two numbers. 2x2x2x2x2. The answer is 32 combinations.
106 or a million.
Just one. In a combination, the order of the digits does not matter.