Hello Everyone, There is one interesting exercise in which it is asked to solve the following problem: In how many ways can we distribute 20 candies among 6 children so that the youngest gets at most 2 candies? This is my version of the solution: Case 1: youngest child gets no candies, thus we have to distribute the 20 candies among 5 children, so we have C(20+5-1, 5) = 42 504 Case 2: youngest child gets one candy, thus we give him one candy, and then we have 19 candies to distribute between the other 5 children: C(19+5-1, 5) = 33 649 Case 3: youngest child gets two candies, so we give him two, and then we have 18 candies to distribute among the other 5 children: C(18+5-1, 5) = 26 334 And the total comes to 42 504 + 26 334 + 33 649 = 102 487 ways to distribute the candies. This is the instructor's version of the solution, whose answer matches the answer in the study guide: You have x1 + x2 + x3 + x4 + x5 + x6 = 20 Nw we will discuss x1, consider y1=x1-i where i is 0, 1, or 2.. So for i=0 we still have x1=y1 and then we still have x1 + x2 + x3 + x4 + x5 + x6 = 20 now I think we can consider x1 as x1=0 is a "choice". To understand my point if you consider x1=1 then using you thinking we will get to x2 + x3 + x4 + x5 + x6 =19 and the answer will be C(19+5-1,19)=C(23,19) however. If you consider i=1 so x1=y1+1 and then the answer will be C(19+6-1,16)=C(24,19), so 6 is always kept for all cases. My question is: which solution is correct?