10 < 1

We can solve this question by taking cases.

First case in which 9 is in the unit's place then 8 number can be placed in one's place.

Similarly when 8 is there in unit place then 7 digits can be placed in ten's place . This series will go till there is 2 in unit place and only 1 can take ten's place.

So total number of numbers are 1+2+3+4………+8 = 36

A combinatorics-less solution,

1x -> the tens digit is 1, therefore there are 8 solutions (1 < 2, 3, ..., 9), 2x -> the tens digit is 2, therefore there are 7 solutions (1 < 2 < 3, 4, ..., 9)

$8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36$

Since the tenth digit is smaller than the unit digit, we are left with {1,2,3,4,5,6,7,8,9} to pick for either tenth or unit place of the digit. If we have to choose two digits out of those 9 digits and order them in strictly descending order,

We can do that in * 9C2 ways * = (9*8)/2 = 36 ways

Alternatively, you can count by each combination: To make sure the number is bigger then the tenth digit:

when the tenth digit is 1, you can pick the unit digit in 8C1 ways (from: 2,3,4,5,6,7,8,9) = 8 ways

when the tenth digit 2, you can pick the unit digit in 7C1 ways (from: 3,4,5,6,7,8,9) = 7 ways

when the tenth digit 3, you can pick the unit digit in 6C1 ways = 6 ways

...

...

when the tenth digit is 8, you can pick the unit digit only in 1 way . i.e.: 89

so, the total no. of digits you can form where the tenth digit is smaller than the unit digit is: 8+ 7+ 6 + 5+ 4+ 3+ 2+ 1 = (8*9)/2 = 36

We can be sure that for every ten numbers (starting 10), the count of the numbers with the tens being higher or equal to the ones place is: 1 plus the tens number itself (e.g. for range 10 - 19, that would be 10 and 11, for range 20-29, that would be 20 until 22).

Therefore, we can safely say that for every range there are 10 - (n +1) or 9 - n numbers whose tens number is lower than their units. Compute for the summation and you get 36.

OR

Since we are terribly sure that there is an arithmetic pattern for the count of disqualified numbers (the numbers whose tens digits are greater/equal to their units) per range, then we can just use the arithmetic series to compute for the total count of these numbers and subtract it from the total count of 2-digit numbers.

Hence, 10n - {(n/2)[(2(total count of disqualified numbers from 10 to 19)) + n-1]}

= 10(9) - {(9/2)[(2(2)) + 9-1]}

= 10(9) - ((9 x 12)/2)

= 90 - 54

= 36

The two digit numbers are :

10, 11, 12, 13, ............ , 97, 98, 99

The required numbers are :

(12, 13. ......., 19), (23, 24, ......., 29), (34, 35, 36, ......39), ...., (78, 79), (89)

8 + 7 + 6 + ....... + 3 + 2 + 1 = (8/2)(8 + 1) = 36

A little long winded but I was searching for the pattern. Take numbers 10 to 19 and lose 10 &11, that leaves 8. Take numbers 20 to 29 lose 20,21,22 leaves 7. This gave me the pattern 8+7+6+5+4+3+2+1=36

1 + 2 + 3 + 4 + ...+ 8 = 9*8/2 =36