Only one.
You can make 5 combinations of 1 number, 10 combinations of 2 numbers, 10 combinations of 3 numbers, 5 combinations of 4 numbers, and 1 combinations of 5 number. 31 in all.
To calculate the number of combinations with three numbers, you would use the formula for combinations, which is nCr = n! / r!(n-r)!. In this case, n is the total number of numbers you have to choose from, and r is the number of numbers you are choosing. So, if you have three numbers to choose from, there would be 3C3 = 3! / 3!(3-3)! = 6 / (6*0!) = 6 / 6 = 1 combination.
No there actually more combinations that we can make with numbers than letters. That's not actually true, since there are more letters than numerals, but every combination of numerals is a number and there are an infinite number of them, whereas, not all combinations of letters actually make words--there is only a finite number of words.
You could make 10*10*10*26*26*26 combinations, or 17576000 combinations.
As a product of its prime factors: 2*5*13 = 130
You can make 5 combinations of 1 number, 10 combinations of 2 numbers, 10 combinations of 3 numbers, 5 combinations of 4 numbers, and 1 combinations of 5 number. 31 in all.
14 * * * * * Wrong! There are 15. 4 combinations of 1 number, 6 combinations of 2 number, 4 combinations of 3 numbers, and 1 combination of 4 numbers.
Assuming that the six numbers are different, the answer is 15.
35
9000
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.
The rearrangement of 5 figure numbers will be 5x4x3x2x1 which is 120 combinations, when you don't repeat a number.
None. You do not have enough numbers to make even one combination.
To calculate the number of combinations with three numbers, you would use the formula for combinations, which is nCr = n! / r!(n-r)!. In this case, n is the total number of numbers you have to choose from, and r is the number of numbers you are choosing. So, if you have three numbers to choose from, there would be 3C3 = 3! / 3!(3-3)! = 6 / (6*0!) = 6 / 6 = 1 combination.
Using the combination fuction, chose three numbers from 45 numbers. The answer is 14,190.
The number of combinations you can make with the digits 1234567890 depends on how many digits you want to use and whether repetition is allowed. If you use all 10 digits without repetition, there are 10! (10 factorial) combinations, which equals 3,628,800. If you are choosing a specific number of digits (for example, 3), the number of combinations would be calculated using permutations or combinations based on the rules you set.
Assuming you are treating each number as a number and not as an individual unit, the numbers you can make from these digits are 899, 989 and 998.