There are tokens numbered from 1 to 100 In how many ways can the tokens be drawn simultaneously so that the sum of two tokens exceed 100?

Answer 1

Drawing two tokens simultaneously is the same as drawing the first and then the second WITHOUT replacing the first. So let the first and second numbers be m and n, and you want m+n > 100. The solution needs to be broken down into two subsets: Subset 1: m <= 50 Then n = 101 - m and n > 50 so there is no possibility of n = m Subset 2: m > 50 Now there are 101 - m ways, but in each case m and n can be equal. So, in fact, there are 101 - m - 1 ways. Combining the two stages, In general, there are 101 - m ways but from subset 2, 50 of these are to be rejected (n=m). Sum this over m = 1 to 100 [My inability to use the sigma notation is causing me some difficulty here.] = 101*100 - (1+2+...+100) - 50 = 101*100 - 101*50 - 50 = 101*50 - 50 = 5050 - 50 = 5000