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If there are no restrictions on the 'combinations' then there are ten choices for the first digit and ten for the second: 10 x 10. (This implies possibilities such as 22 and 77.) If the digits must be different in each combination then the number of combinations taking two at a time from ten is C(10,2) = 10!/( 2! ( 10 - 2 )! ) = 45.
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.
35
9000
None. You do not have enough numbers to make even one combination.
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.
If there are no restrictions on the 'combinations' then there are ten choices for the first digit and ten for the second: 10 x 10. (This implies possibilities such as 22 and 77.) If the digits must be different in each combination then the number of combinations taking two at a time from ten is C(10,2) = 10!/( 2! ( 10 - 2 )! ) = 45.
5435 441521 14241553 514152153 457745641 543646313 IDK TO HarD!
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.
Only one.
Assuming that the six numbers are different, the answer is 15.
35
9000
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.
u culd make 80
Using the combination fuction, chose three numbers from 45 numbers. The answer is 14,190.