How many integers from 1 to 50 are multiples of 2 or 3 but not both?įrom 1 to 100, there are $50/2 = 25$ numbers which are multiples of 2. Examplesįrom a set S =|A_1 \cap \dots \cap A_2|$ In other words a Permutation is an ordered Combination of elements. PermutationsĪ permutation is an arrangement of some elements in which order matters. Hence from X to Z he can go in $5 \times 9 = 45$ ways (Rule of Product). Thereafter, he can go Y to Z in $4 + 5 = 9$ ways (Rule of Sum). Solution − From X to Y, he can go in $3 + 2 = 5$ ways (Rule of Sum). How many ways are there to go from X to Z? From there, he can either choose 4 bus routes or 5 train routes to reach Z. He may go X to Y by either 3 bus routes or 2 train routes. From his home X he has to first reach Y and then Y to Z. Question − A boy lives at X and wants to go to School at Z. Mathematically, if a task B arrives after a task A, then $|A \times B| = |A|\times|B|$ The Rule of Product − If a sequence of tasks $T_1, T_2, \dots, T_m$ can be done in $w_1, w_2, \dots w_m$ ways respectively and every task arrives after the occurrence of the previous task, then there are $w_1 \times w_2 \times \dots \times w_m$ ways to perform the tasks. $A \cap B = \emptyset$), then mathematically $|A \cup B| = |A| + |B|$ If we consider two tasks A and B which are disjoint (i.e. The Rule of Sum − If a sequence of tasks $T_1, T_2, \dots, T_m$ can be done in $w_1, w_2, \dots w_m$ ways respectively (the condition is that no tasks can be performed simultaneously), then the number of ways to do one of these tasks is $w_1 + w_2 + \dots +w_m$. The Rule of Sum and Rule of Product are used to decompose difficult counting problems into simple problems. Counting mainly encompasses fundamental counting rule, the permutation rule, and the combination rule. For solving these problems, mathematical theory of counting are used. For instance, in how many ways can a panel of judges comprising of 6 men and 4 women be chosen from among 50 men and 38 women? How many different 10 lettered PAN numbers can be generated such that the first five letters are capital alphabets, the next four are digits and the last is again a capital letter. Then S = A c ∩ B c∩ C c since each element of S is not divisible by 3, 5, or 7.In daily lives, many a times one needs to find out the number of all possible outcomes for a series of events. Let C be the subset of integer which is divisible by 7 Let B be the subset of integer which is divisible by 5 Solution: Let A be the subset of integer which is divisible by 3 Then |U|= 1000 Find |S| where S is the set of such integer which is not divisible by 3, 5 or 7? Then the number m of the element which do not appear in any subset A 1,A 2.A r of U.Įxample: Let U be the set of positive integer not exceeding 1000. Let A 1,A 2.A r be the subset of Universal set U. So, according to the pigeonhole principle, there must be at least two people assigned to the same month. Solution: We assigned each person the month of the year on which he was born. Solution: Here n = 12 months are the PigeonholesĮxample2: Show that at least two people must have their birthday in the same month if 13 people are assembled in a room. 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.Įxample1: Find the minimum number of students in a class to be sure that three of them are born in the same month. If n pigeonholes are occupied by n+1 or more pigeons, then at least one pigeonhole is occupied by greater than one pigeon.
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