Question

The 8 digits 0 through 7 are to be arranged in a 2x4 table (a table with 2 rows and 4 columns). How many ways are there to do this if for each column, the entry in row 1 must be less than the entry in row 2?

Answer 1

You have 7 ways of filling the 1,1 number, and 1 way if you chose 7, 2 if you chose 6... 7 if you chose 1 to fill the 1,2 number, 7*(1+2+...+7)=7*28
So you have 196 ways of filling the first 2 places.
Then you have 5 ways of filling the 2,1 number, and 1+2+3+4+5 of filling the 2,2 number.
Then 3 ways of filling 3,1 and 1+2+3 of filling the the 3,2
Then 1 way of filling the last one and 1 way of filling it.
So you have 7*28*5*15*3*6*1*1=196*75*18=264600
But we counted more than 1 time some cases.
24 times to be exact (4!) I think.
So there are 11025 ways of filling the table.

I'm not 100% sure if this is right, if I were I'd try to help you in other way instead of trying to solve the problem :p