I had a dream last night involving — (?) well I am not really sure, except that it left me wondering if there is a simple proof (if indeed it is true) that there must be a common factor of
m choose i = m!/(i! (m-i)!)
m choose j = m!/(j! (m-j)!)
for all counting numbers i,j,m with 1 < i,j < m Another way to state this same thing is: any pair of entries, on any row of Pascal’s triangle (except for the 1’s on the edges) will have a common factor. With facts of this sort, often there is a clever way to cast things in terms of counting something a couple of different ways which makes things clear.