First > second

$10 = 1$

$20, 21 = 2$

$30, 31, 32 = 3$

Since we are taking two digit numbers, we go up to $99$ and get $9$ such numbers.

Thus $1+2+3+4+5+6+7+8+9 = 45$

The answer is $T_9 = \frac{1}{2}(9) (9+1) = \boxed{45}$ This is because there are $n$ satisfactory numbers when the tens digit is equal to $n.$

Python:

1
2
3
4
5
6

count = 0
for n in xrange(10,100):
    s = str(n)
    if int(s[0]) > int(s[1]):
        count += 1
print "Answer:", count

There is just 1 digit lesser than 1 i.e. 0. So, the 2-digit number 10 follows this criteria.

There are 2 digits less than 2 i.e. 1 and 0. So, 20 and 21 follow the criteria.

Hence, there is 1 number starting with 1, 2 numbers starting with 2 and so on. So, sum of such numbers =1+2+3+4+5+6+7+8+9=45

for each n from 1 to 9 there are n choices for y, the second digit. e.g. 9, the second digit can be 0,1,2,3,4,5,6,7,8 = 9 choices.

so 1+2+3+4+5+6+7+8 +9 = 45