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.
You can make 6 combinations with 3 numbers. They are: 123 213 312 132 231 321 * * * * * NO! Those are permutations! In combitorials, the order does not matter so that the combination 123 is the same as the combination 132 etc. So all of the above comprise just 1 combination. With three numbers you can have 1 combination of three numbers (as discussed above), 3 combinations of 2 numbers (12, 13 and 23) 3 combinations of 1 number (1, 2 and 3) In all, with n numbers you can have 2n - 1 combinations. Or, if you allow the null combination (that consisting of no numbers) you have 2n combinations.
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
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.
9000
The rearrangement of 5 figure numbers will be 5x4x3x2x1 which is 120 combinations, when you don't repeat a number.
To find the number of different combinations of the numbers 1 to 10, we can consider the combinations of choosing any subset of these numbers. The total number of combinations for a set of ( n ) elements is given by ( 2^n ) (including the empty set). For the numbers 1 to 10, ( n = 10 ), so the total number of combinations is ( 2^{10} = 1024 ). This includes all subsets, from the empty set to the full set of numbers.
None. You do not have enough numbers to make even one combination.
You can make 6 combinations with 3 numbers. They are: 123 213 312 132 231 321 * * * * * NO! Those are permutations! In combitorials, the order does not matter so that the combination 123 is the same as the combination 132 etc. So all of the above comprise just 1 combination. With three numbers you can have 1 combination of three numbers (as discussed above), 3 combinations of 2 numbers (12, 13 and 23) 3 combinations of 1 number (1, 2 and 3) In all, with n numbers you can have 2n - 1 combinations. Or, if you allow the null combination (that consisting of no numbers) you have 2n combinations.
Using the combination fuction, chose three numbers from 45 numbers. The answer is 14,190.
First you need to find you first number. As there are five numbers, the first number can be any of the five. So you have five combinations (5). Next you take your second number, as you've already used one you can only choose one of four numbers (5 x 4). You third number, you can only choose one of three numbers (5 x 4 x 3), you fourth, one of two (5 x 4 x 3 x 2) and finally you are left with one number, so you only have one choice. So we have a combination of 5 x 4 x 3 x 2 x 1 (or 5!) to find the number of different 5 number combinations you can make with 1, 2, 3, 4 and 5. Answer: 120.