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?

Question

zarkon · Answer

looks like 113,400

ybarrap · Answer

row2,col1 has 10 possible entries
row1,col1 has 9 possible entries

row2,col2 has 8 possible entries
row1,col2 has 7 possible entries

continuing in this fashion:

row2 has 10*8*6*4*2 possible values
row 1 has 9*7*5*3*1 possible values

Total possibilities are 10! that meet the specified condition.

anonymous · Answer

Zarkon can you please explain how you did this one as well? you are so smart!

zarkon · Answer

it is really easy....

zarkon · Answer

there are 10! ways to arrange the numbers ....

zarkon · Answer

there are 2^5 ways to permute the rows only one of which will make the first column smaller than the 2nd column..to the number of ways to do this is 

$$\frac{10!}{2^5}$$

anonymous · Answer

Thank you!!