5435
441521
14241553
514152153
457745641
543646313
IDK TO
HarD!
If repeats are allowed than an infinite number of combinations is possible.
2469135780
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.
1.52415788 × 1018
1234567890 - 987654321 = 246913569
If repeats are allowed than an infinite number of combinations is possible.
The answer is 10C3 = 10*9*8/(3*2*1) = 120 combinations.
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.
1234567890
2469135780
zero
1.52415788 × 1018
1234567890 - 987654321 = 246913569
1234567890-1234567856 = 34
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.
1 plus 1234567890 is 1234567891
They are all single digit numbers.