problem sum Home › Forums › Math Olympiad, I.S.I., C.M.I. Entrance › Combinatorics › problem sum This topic contains 2 replies, has 3 voices, and was last updated by A 1 month ago. Viewing 3 posts - 1 through 3 (of 3 total) Author Posts October 12, 2019 at 6:56 pm #39189 amit poddarParticipant Problem 19. How many ten-digit numbers have the sum of their digits equal to a) 2; b) 3; c) 4? October 15, 2019 at 8:52 am #39327 Jatin Kr DeyParticipant Hello Amit, your question is not clear to us. Can you please send any photo of it? October 15, 2019 at 4:39 pm #39347 AParticipant Firstly, I assume that the question you asking is to find the number of 10-digit numbers whose digit sum is 2,3, and 4 respectively, in separate cases. Subdivision (a) :- Digit sum is 2 We have that 2 = 0+2 or 1+1. So we can either have a 10-digit number composed of only one 2 and nine 0’s, or a 10-digit number composed of two 1’s and eight 0’s. In the first case, there is only one such number, that is 2000000000, and nothing else, because 2 can only be in the first digit’s place. In the latter, we have a 1 to be in the first digit’s place and another 1 has 9 other places. So totally, in this case we have 1 x 9 = 9 such numbers, and in the first case we have $1$ number, so totally there are 10 such numbers. Similarly, you can follow this principle for the other two subdivisions too. Hope this helps. This reply was modified 1 month ago by A. Author Posts Viewing 3 posts - 1 through 3 (of 3 total) You must be logged in to reply to this topic.