Lecture 32 Discrete Mathematics (MTH202) PDF

Discrete Mathematics (MTH202) LECTURE 32 ORDERED AND UNORDERED PARTITIONS PERMUTATIONS WITH REPETITIONS K-SELECTIONS: k-selections are similar to k-combinations in that the order in which the elements are selected does...

Discrete Mathematics (MTH202) LECTURE 32 ORDERED AND UNORDERED PARTITIONS PERMUTATIONS WITH REPETITIONS K-SELECTIONS: k-selections are similar to k-combinations in that the order in which the elements are selected does not matter, but in this case repetitions can occur. DEFINITION: A k-selection of a set of n elements is a choice of k elements taken from a set of n elements such that the order of elements does not matter and elements can be repeated. REMARK: 1. k-selections are also called k-combinations with repetition allowed or multisets of size k. 2. With k-selections of a set of n elements repetition of elements is allowed. Therefore k need not to be less than or equal to n. THEOREM: The number of k-selections that can be selected from a set of n elements is k+n-1 C(k+n−1, k) or c k EXERCISE: A camera shop stocks ten different types of batteries. (a) How many ways can a total inventory of 30 batteries be distributed among the ten different types? (b) Assuming that one of the types of batteries is A76, how many ways can a total inventory of 30 batteries be distributed among the 10 different types if the inventory must include at least four A76 batteries? SOLUTION: (a) k = 30 n = 10 The required number is C(30 + 10 – 1, 30)= C(39, 30) 39! = (39 − 30)!30! = 211915132 (b) k = 26 n = 10 The required number is C(26 + 10 – 1, 26) = C(35, 26) = 35! (35 − 26)!26! Page 1 of 6 © Copyright Virtual University of Pakistan Discrete Mathematics (MTH202) = 70607460 WHICH FORMULA TO USE? ORDER ORDER DOES MATTERS NOT MATTER REPETITION k-sample k-selection ALLOWED nk C(n+k-1, k) REPETITION NOT k-permutation k-combination ALLOWED P(n, k) C(n, k) ORDERED AND UNORDERED PARTITIONS: An unordered partition of a finite set S is a collection [A1, A2, …, Ak] of disjoint (nonempty) subsets of S (called cells) whose union is S. The partition is ordered if the order of the cells in the list counts. EXAMPLE: Let S = {1, 2, 3, …, 7} The collections P1 = [{1,2}, {3,4,5}, {6,7}] And P2 = [{6,7}, {3,4,5}, {1,2}] determine the same partition of S but are distinct ordered partitions. EXAMPLE: Suppose a box B contains seven marbles numbered 1 through 7. Find the number m of ways of drawing from B firstly two marbles, then three marbles and lastly the remaining two marbles. SOLUTION: The number of ways of drawing 2 marbles from 7 = C(7, 2) Following this, there are five marbles left in B. The number of ways of drawing 3 marbles from 5 = C(5, 3) Finally, there are two marbles left in B. The number of way of drawing 2 marbles from 2 = C(2, 2) Thus  7   5  2  m =      2   3  2  7! 5! 2! = ⋅ ⋅ 2!5! 2!3! 2!0! 7! = = 210 2!3!2! Page 2 of 6 © Copyright Virtual University of Pakistan Discrete Mathematics (MTH202) THEOREM: Let S contain n elements and let n1, n2, …, nk be positive integers with n1+n2+…+nk = n. Then there exist n! n1 ! n2 ! n3 !L nk ! different ordered partitions of S of the form [A1, A2, …, Ak], where A1 contains n1 elements A2 contains n2 elements A3 contains n3 elements ……………………….. Ak contains nk elements REMARK: To find the number of unordered partitions, we have to count the ordered partitions and then divide it by suitable number to erase the order in partitions. EXERCISE: Find the number m of ways that nine toys can be divided among four children if the youngest child is to receive three toys and each of the others two toys. SOLUTION: We find the number m of ordered partitions of the nine toys into four cells containing 3, 2, 2 and 2 toys respectively. Hence 9! m = 3!2!2!2! = 2520 EXERCISE: How many ways can 12 students be divided into 3 groups with 4 students in each group so that (i) one group studies English, one History and one Mathematics. (ii) all the groups study Mathematics. SOLUTION: (i) Since each group studies a different subject, so we seek the number of ordered partitions of the 12 students into cells containing 4 students each. Hence there are 12! such partitions = 34,650 4!4!4! (ii) When all the groups study the same subject, then order doesn’t matter. Now each partition {G1, G2, G3} of the students can be arranged in 3! ways as an ordered partition, hence there are 12! 1 × 4!4!4! 3! Page 3 of 6 © Copyright Virtual University of Pakistan Discrete Mathematics (MTH202) unordered partitions. EXERCISE: How many ways can 8 students be divided into two teams containing (i) five and three students respectively. (ii) four students each. SOLUTION: (i) The two teams (cells) contain different number of students; so the number of unordered partitions equals the number of ordered partitions, which is 8! = 56 5!3! (ii) Since the teams are not labeled, so we have to find the number of unordered partitions of 8 students in groups of 4. 8! Firstly, note, there are = 70 ordered partitions into two cells containing four students each. 4!4! Since each unordered partition determine 2! = 2 ordered partitions, there are 70 = 35 2 unordered partitions EXERCISE: Find the number m of ways that a class X with ten students can be partitions into four teams A1, A2, B1 and B2 where A1 and A2 contain two students each and B1 and B2 contain three students each. SOLUTION: There are 10! = 25, 200 ordered partitions of X into four cells 2!2!3!3! containing 2, 2, 3 and 3 students respectively. However, each unordered partition [A1, A2, B1, B2] of X determines 2!⋅2! = 4 ordered partitions of X. Thus, 25, 200 m= = 6300 4 EXERCISE: Suppose 20 people are divided in 6 (numbered) committees so that 3 people each serve on committee C1 and C2, 4 people each on committees C3 and C4, 2 people on committee C5 and 4 people on committee C6. How many possible arrangements are there? SOLUTION: We are asked to count labeled group - the committee numbers labeled the group.So this is a problem of ordered partition. Now, the number of ordered partitions of 20 people into the specified committees is 20! = 2444321880000 3!3!4!4!2!4! Page 4 of 6 © Copyright Virtual University of Pakistan Discrete Mathematics (MTH202) EXERCISE: If 20 people are divided into teams of size 3, 3, 4, 4, 2, 4, find the number of possible arrangements. SOLUTION: Here, we are asked to count unlabeled groups. Accordingly, this is the case of ordered partitions. Now number of ordered partitions = 20! 1 × 3!3!4!4!2!4! 3!2! = 203693490000 GENERALIZED PERMUTATION or PERMUTATIONS WITH REPETITIONS: The number of permutations of n elements of which n1 are alike, n2 are alike, …, nk are alike is n! n1 ! n2 !L nk ! REMARK: n! The number n ! n !L n ! is often called a multinomial coefficient, and is denoted by the 1 2 k symbol.  n   n , n ,L , n   1 2 k  EXERCISE: Find the number of distinct permutations that can be formed using the letters of the word “BENZENE”. SOLUTION: The word “BENZENE” contains seven letters of which three are alike (the 3 E’s) and two are alike (the 2 N’s) Hence, the number of distinct permutations are: 7! = 420 3!2! EXERCISE: How many different signals each consisting of six flags hung in a vertical line, can be formed from four identical red flags and two identical blue flags? SOLUTION: We seek the number of permutations of 6 elements of which 4 are alike and 2 are alike. 6! There are = 15 different signals. 4!2! EXERCISE: (i)Find the number of “words” that can be formed of the letters of the word ELEVEN. (ii)Find, if the words are to begin with L. (iii)Find, if the words are to begin and end in E. (iv)Find, if the words are to begin with E and end in N. SOLUTION: Page 5 of 6 © Copyright Virtual University of Pakistan Discrete Mathematics (MTH202) (i)There are six letters of which three are E; hence required number of “words” are 6! = 120 3! (ii)If the first letter is L, then there are five positions left to fill where three are E; hence required number of words are 5! = 20 3! (iii)If the words are to begin and end in E, then there are only four positions to fill with four distinct letters. Hence required number of words = 4! = 24 (iv)If the words are to begin with E and end in N, then there are four positions left to fill where two are E. 4! Hence required number of words = = 12 2! EXERCISE: (i)Find the number of permutations that can be formed from all the letters of the word BASEBALL (ii)Find, if the two B’s are to be next to each other. (iii)Find, if the words are to begin and end in a vowel. SOLUTION: (i)There are eight letter of which two are B, two are A, and two are L. Thus, 8! Number of permutations = 2!2!2! = 5040 (ii)Consider the two B’s as one letter. Then there are seven letters of which two are A and two are L. Hence, 7! Number of permutations = 2!2! = 1260 (iii)There are three possibilities, the words begin and end in A, the words begin in A and end in E, or the words begin in E and end in A. In each case there are six positions left to fill where two are B and two are L. Hence, 6! Number of permutations =3 =540 2!2! Page 6 of 6 © Copyright Virtual University of Pakistan