combinatorics ace mock

Question

The number of integers between 1 and 1000 inclusive have a sum of digits equal to 10 is_______

Answer 1

The number of solutions to a + b + c = 10 in non-negative integers is 12 * 11 / 2 = 66. From this list we must exclude the solutions in which one of the variables is 10. There are three such solutions, so 63 solutions remain.

Comments

My question was, here we just need to consider even 2 digits (Eg: 9+1=10 ) to till 9 digits (1+1+1.....+2=10)

all those cases also should be considered right ?

Correct me if i'm wrong!

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Can't get what you asking but question say to only consider till 1000 so why will I take 5 digits into consideration and I have placed a box for every place place,  for unit place for one for 10's place and ones for 1000 place, 1000 does not sum up to 10 so we have to consider the possibility till 0-999

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Yes got it i misunderstood the question now got cleared. Thanks