Four is not more than two ?

Evidently any no. so formed of 4 digits contains 1#- only one digit (like 1111,2222,......) and there are 9 numbers. 2#- two digits (A) if 0 is one of the two,then the one more can be anyone of the nine, and these two digits can be arranged in 9C1[ 3C1+3C2+3C3] =63 (B) if 0 is not one of them,then two of the digits have to be selected from 9, and these two can be arranged in 9C2[4C2+4C2+4C3] =504

HENCE TOTAL NO. OF REQUIRED NUMBERS=576

X : number of distinct digit in a 4 digit number. We are looking for |X <= 2| = |X = 0| + |X = 2|; X = 1 has no sense (to me at least)

|X = 0| is obviously 10 (or 1c10 if you like). |X = 2|, let's distinguish two case : -both of the two distinct digit are repeated one time each two form a 4 digit number, that's make : 2c10 * 2c2 * (4!/2!2!) possibilities -one of the two distinct digit are repeated 3 time, we then have : 2c10 * 1c2 * 4!/3! possibilities.

by adding the possibilities, we have 640 cases. since we treated the number that start by an 0 - aka 3 digit number, which represent 1/10 of the cases, the answer is then : 640 * 9/10 = 576.