How many integers between 1 and 10^81 are such that the sum of their digits is 2? A) 3160 B) 3240 C) 3241 D) 3320 E) 3321

IMO (E) Sum of all the digits will be 2 only when either have only if one appears twice in a number and rest should be zero or there is 1 two present in the number and rest should be zero. for example, Up to 4 digits:- Case 1: 1100 1001 1010 0011 Case 2: 2000 0200 0020 0002 Since the upper limit is 10^81 therefore there will 81 places where we can arrange the combination of 2 ones and zeros or 1 two and zeros. For case 1: The total number of cases will be: 81C1 x 80C80 = 81 For case 2: 81C2 x 79C79 = 3240 Therefore total cases: 3240 + 81 = 3321 Choose (E)

Nice work, Daisy. The answer is E I'll post my solution as well: Begin by noting that, if the sum is 2, then there are two cases: Case 1: The number has two 1’s and all remaining digits are 0 Case 2: The number has one 2 and all remaining digits are 0 If a number is less than 10^81, then that number has 81 digits or fewer. For this question, we’ll say that all numbers less than 10^81 have EXACTLY 81 digits, however some of those numbers have leading digits of zero (e.g., 0000…0001001 = 1001 and 0000…00200 = 200). Case 1: In how many ways can we have an 81-digit number such that there are two 1’s and seventy-nine 0’s? Essentially, we have 81 places and we must select 2 of those places to hold the digit “1.” We can accomplish this selection in 81C2 (3240) ways. So, there are 3240 numbers consisting of two 1’s Case 2: In how many ways can we have an 81-digit number such that there is one 2 and the remaining digits are 0? We have 81 places and we must select 1 place to hold the 2 (the remaining places will be filled with 0’s). There are 81 places and we must select 1 place. We can accomplish this in 81 ways. Adding the results from Cases 1 and 2, we get a total of 3321 numbers.