jmc

In addition to two-digit numbers, consider the one-digit numbers $01$ through $09.$ Of these $99$ numbers, $9$ have repeated digits $01,$ $02,$ $03,$ $\dots ,$ $99,$ namely: $11,$ $22,$ $33,$ $\dots ,$ and $99.$ Of the remaining $90$ numbers, each one has a unique counterpart formed by swapping their tens and units digit. Thus among these, $45$ have a tens digit greater than the units digit. (The other $45$ have a tens digit smaller than the units digit.) The required answer is $\boxed{45}.$