Puzzle #232 Solution: Hugs and Kisses Spoiler Inside: Solution to Puzzle SelectShow> Before the limitation of at least two, the possible combinations are zero to three X’s and zero to two O’s. These are independent so the possibilities are multiplied to give twelve possibilities. Zero of each is not allowed (eliminating one possibility). One of each is not allowed (eliminating O and X). This leaves nine possibilities. They are OO, OX, OOX, XX, OXX, OOXX, XXX, OXXX, and OOXXX.