The Pigeonhole PrincipleIf n pigeonholes are occupied by n+1 or more pigeons, then at least one pigeonhole is occupied by greater than one pigeon. Generalized pigeonhole principle is: - If n pigeonholes are occupied by kn+1 or more pigeons, where k is a positive integer, then at least one pigeonhole is occupied by k+1 or more pigeons. Example1: Find the minimum number of students in a class to be sure that three of them are born in the same month. Solution: Here n = 12 months are the Pigeonholes Example2: Show that at least two people must have their birthday in the same month if 13 people are assembled in a room. Solution: We assigned each person the month of the year on which he was born. Since there are 12 months in a year. So, according to the pigeonhole principle, there must be at least two people assigned to the same month. Inclusion-Exclusion Principle:Let A_{1},A_{2}......A_{r} be the subset of Universal set U. Then the number m of the element which do not appear in any subset A_{1},A_{2}......A_{r} of U. Example: Let U be the set of positive integer not exceeding 1000. Then |U|= 1000 Find |S| where S is the set of such integer which is not divisible by 3, 5 or 7? Solution: Let A be the subset of integer which is divisible by 3 Then S = A^{c} ∩ B^{c}∩ C^{c} since each element of S is not divisible by 3, 5, or 7. By Integer division, |A|= 1000/3 = 333 Thus by Inclusion-Exclusion Principle |S|=1000-(333+200+142)+(66+47+28)-9 Next TopicRecurrence Relations |