Question

The 10 digits 0 through 9 are to be arranged in a 2x5 table( a table with 2 rows and 5 columns). How many ways are there to do this if for each column, the entry in row 1 must be less then the entry in row 2?

Answer 1

This just might be correct:

Let's denote the numbers by (x,x). So (2,1) is the number in the second row, first column. There 10! total combinations. Half of those have (1,1) less than (2,1). So we're left with 10!/2 combinations. Half those again have (1,2) less than (2,2) etc. So I'm guessing there are a total of 10!/2^5 combinations.