Assuming all 9 numbers are one digit and different
1st number can be any of the 9 numbers
2nd any of the remaining 8 numbers
3rd any of the remaining 7 numbers
4th any of the remaining 6 numbers
5th any of the remaining 5 numbers
So total is 9 x 8 x 7 x 6 x 5 = 15,120 combinations
* * * * *
That is the number of permutations, not combinations.
In the above method, 12345, 12354, 12435, etc would all be counted as different even though they are all the SAMEcombination of numbers. Thus there are 120
The correct answer is 9C5 = 9*8*7*6*5/(5*4*3*2*1) = 126 combinations.
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.
9000
35
None. You do not have enough numbers to make even one combination.
There are a huge number of combinations of 5 numbers when using the numbers 0 through 10. There are 10 to the 5th power combinations of these numbers.
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.
Only one.
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.
9000
252 combinations, :)
35
None. You do not have enough numbers to make even one combination.
The rearrangement of 5 figure numbers will be 5x4x3x2x1 which is 120 combinations, when you don't repeat a number.
There are 8,592,039,666 combinations of 6 numbers out of 138 numbers, like the numbers from 1 to 138.
There are a huge number of combinations of 5 numbers when using the numbers 0 through 10. There are 10 to the 5th power combinations of these numbers.
There are infinitely many numbers and so infinitely many possible combinations.