Question 563661
If the 1st number is 1, the 2nd can be 2-10
If the 1st number is 2, the 2nd can be 3-10
If the 1st number is 3, the 2nd can be 4-10
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This progression is
{{{ 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45 }}}
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A formula would be if you can choose numbers 
from 1 to n, then the possible arrangements of 2 numbers
the 1st being smaller than the 2nd
is  (n-1) + (n-2) + (n-3) + . . . + 1
This sum is {{{ (1/2)*n*(n-1) }}}
check:
for 1 to 10, {{{ (1/2)*10*9 = 90/2 }}}
{{{ 90/2 = 45 }}} 
OK