There are 10C2=45 ways to assign pairs (x,y) — note that order is not important here, because x,y together represent the 2 digits allowed in the number. Take special care when either one is 0.

If neither are 0, there are 2^k lucky numbers that are k digits containing those numbers. If x is 0, then it's 2^(k-1). But this double counts the numbers if x=y.

a=10^k (ie. k 9's): # =9x2^(k-1)-9+36x2^k-36 -9k =81x2^(k-1)-45-9k (*)

We subtract 9k because that is precisely the number of lucky numbers containing only 1 digit

Thereafter, the only necessary consideration left is the last segment of lucky numbers from a to n, all of which are k digits long.

For example, in the given test case, n=123, ans=113.

a=100=10^2=02 x 9's (as in '99') => k=3

By (*), k=2: #=99; k=3: #=252

There are 99 lucky numbers between 1 and 99 inclusive.

Between a=100 <= n=123, there are 3+9+2=14 more lucky numbers, specifically as follows:

• 100,101,110
• 111 — 119
• 121, 122