- #1

sylar

- 11

- 0

Here for each n, we must find the possibility of having an intersection set of elements, multiply this probability by n, and then sum up the products we obtained.

Take the case when n=3. Then there are 8 different possibilities for choosing A, and also for B. Thus, there are 64 different possible selections of A int. B. We must find the possibility of having the set A int. B with n elements, where n=0,1,2,3, and this seems very complicated. So, is there a better approach for this (general) problem? Thanks!