about... 25 just take 5 numbers times 5 numbers
6 2-digit numbers (13/15/35/53/51/31)
55----------------------------------------------------------------------------------------------From Rafaelrz.You can make, 5! = 120, five digit numbers using 1,2,3,4 and 6.
106------------------------------------------------------------------------------------------------From Rafaelrz.You can make, 4P3 = 4!/(4-3)! = 24, different 3 digit numbers using 1, 2, 3 and 4.
256!
Assuming that the six numbers are different, the answer is 15.
48
about... 25 just take 5 numbers times 5 numbers
120
Their is 25 combinations
6 2-digit numbers (13/15/35/53/51/31)
If you want 4-digit numbers, there are 24 of them.
55----------------------------------------------------------------------------------------------From Rafaelrz.You can make, 5! = 120, five digit numbers using 1,2,3,4 and 6.
106------------------------------------------------------------------------------------------------From Rafaelrz.You can make, 4P3 = 4!/(4-3)! = 24, different 3 digit numbers using 1, 2, 3 and 4.
256!
couple=2
There are infinitely many rational numbers between any two (different) numbers, no matter how close together they are.