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

Sounds as though the number of possible combinations from 8 digits taken four at a time is
C
8 4

which translates to

8!
-----
4!(8-4)!

which evaluates to

8*7*6*5
-------
4!

which could be simplified further.


Anyone care to discuss how to calculate the number of combinations available for the 2nd row when in every one of the 4 columns every entry in row 1 must be less than the entry in row 2?


Just one example might be as follows:

6 4 3 2
7 5 4 5 is just one possibility for row 2 values, each of which must be greater than the digit immediately above it.