Please log in to comment and participate in discussions.
ans is 118
Academic discussion and doubt solving for: Logarithms, Set Theory, Matrices & Determinants, Permutation & Combination, Probability, Binomial Theorem
Create an account to join our community and participate in discussions.
Three scenarios: 2-digit numbers, 3-digit numbers and 4-digit numbers up to 1999. (1-digit number isn't possible).
2-digit number is fairly easy to count: 19, 91, 28, 82, 37, 73, 46, 64, 55. So, 9 numbers.
3-digit numbers: a+b+c=10. Use 10 'p', and every digit except for the first has to more than equal to 0. So, give one of the 'p' to 'a' (this ensures that a is always greater than 1 and we can take a= 1+a1). So a1+b+c=9. Now take 9 'p' and 2 'i' as the dividers. So, we have to arrange 9 'p' and 2 'i's. Hence, 11C2=55.
4-digit numbers: a+b+c+d=10. a can be 1 only, so b+c+d=9. This is same as the 3-digit case and we get 55 as the result.
So sum of all the results: 55+55+9=119. The answer is 118, I checked with Gemini. This discrepency arises because I counted 1000 in the first case as well. This is because 'a' was already allotted one p, and a1+b+c=9. This also included the possibility where a1 is 9, which makes 'a' 10 (a=1+a1), and 'b', 'c' 0, which isn't valid. So, 119-1=118 solutions.
We must keep track of these things in the real exam...