The number of integers between 1 and 1000000 which have the sum of their digits equal to 18 is

First, we can ignore 1000000 since its sum of digits is not 18. Also, let's add 0's in front of numbers to have 6 digits if the number has less than 6 digits. So our range is going to be from 000001 to 999999.

Number of partition of 18 into 6 whole number is 
(18+6-1)C(6-1)=(23)C(5). Assign these 6 numbers to the 6 digits.

Now, we need to remove the cases where one digit is 10 or bigger. It is very good thing that two digits cannot be 10 or bigger, as the sum would go over 18.

Let one digit has x, where x>=10. The sum of the other 5 digits is 18-x. Since 18-x is partitioned into 5 whole numbers, we have (18-x+4)C(4) possible ways. This should be multiplied by 6 since the big digit can be the first digit, second digit, ..., sixth digit. x is from 10 to 18. So we need to calculate 6((12)C(4) + (11)C(4) + .... + (4)C(4)).

Since (n)C(m)+(n)C(m+1)=(n+1)C(m+1), we get 
(4)C(4) + (5)C(4) + (6)C(4) + (7)C(4) + .... + (12)C(4) 
=((5)C(5) + (5)C(4)) + (6)C(4) + (7)C(4) + .... + (12)C(4) 
=((6)C(5) + (6)C(4)) + (7)C(4) + .... + (12)C(4) 
... 
=(13)C(5). (This shows beauty of mathematics, doesn't it?)

Therefore, the answer is (23)C(5)-6*(13)C(5)=25927.