1. Number of Elements in a Finite Set

2. Notation : The number of the elements of the set A will be denoted by n(A) Examples (1): Let: Let: A = {a,b,c,d} B = {1,8,1000,-9} C = {0} D = { ☺} What is n(A), n(B), n(C), n(D) and Φ ? Solution: n(A) = n(B) = 4 n(C) = n(D) = 1 n(Φ) = 0

3. Number of elements in the Union of Two Finite Sets 1. If A and B are disjoint, then: n(AUB) = n(A) + n(B) 2. in general: n(AUB) = n(A) + n(B) – n(A∩B)

4. Example (2) Let A = {1,3,4,5}, B = {2,6,7,8,9 }and C = {1,2,3,7,8}. Find n(AUB), n(AUC), n(BUC} and show that this is consistent with the given above formula Solution: 1. n(AUB) = n( {1,3,4,5}U {2,6,7,8,9 ) = n( {1,2,3,4,5,6,7,8,9} ) = 9 This is consistent with first formula, since we have:n(A) + n(B) = 4 + 5 = 9 It also consistent with the second formula, since n(A∩B) = n(Φ) = 0, and so n(AUB) = n(A) + n(B) - n(A∩B) = 4 + 5 - 0 = 9 2. n(AUC) = ( {1,3,4,5}U{1,2,3,7,8}) = n ( {1,2,3,4,5,7,8} ) = 7 = This is consistent with the second formula, since, n(A∩B) = n({1,3}) = 2 and so: n(A) + n(B) – n(A∩B) = 4 + 5 – 2 = 7

5. 3. n(BUC) = ({2,6,7,8,9 } U{1,2,3,7,8}) = n ( {1,2,3,6,7,8,9} ) = 7 This is consistent with the second formula, since n(B∩C) = n(B) + n(C) – n(B∩C) = n({2,6,7,8,9 }) + n({1,2,3,7,8}) – n({2,7,8}) = 5 + 5 – 3 = 7

6. Illustration n(AUB) =n ({2,4,6,8 } U{2,3,5,6}) = n ( {2,3,4,5,6,8 ) = 6 This is consistent with the second formula, since n(A∩B) = n(A) + n(B) – n(A∩B) = n({2,4,6,7,8 }) + n({2,3,5,6}) - n({2,7,8}) = 4 + 4 – 2 = 6 A AUB 6 2 5 3 8 4 B

7. Example (3) In this class there are 26 students. If 10 students got a score of no less than 16 in the first exam, 11 students got a score of no less than 16 in the second exam and 9 students got a score of no less than 16 in the both exams. If I want to list the names of the students who got such score in either exam, how many students there will be in that list?

8. Solution Let A = The set of all student with a score of no less than 16 in the first exam. → n(A) = 10 B = The set of all student with a score of no less than 16 in the second exam. → n(B) = 11 Then. A∩B = The set of all student with a score of no less than 16 in both exams. → n(A∩B = 9 AUB = The set of all student with a score of no less than 16 in either exam. This is the set of the students who will be in the list. The number of these students = n (AUB) = n(A) + n(B) – n(A∩B ) = 10 + 11 – 9 =12

9. The Multiplication Principle (Basic Counting Principle) If a task (stage of a procedure) T 1 can be performed in n 1 ways, a task T 2 can be performed in n 2 ways, a task T 3 can be performed in n 3 ways, ………………………………………… a task T m can be performed in n m ways, Then: The number of ways of performing the tasks T 1, T 1, T 1, …. T m in Succession (completing the procedure) is equal to: n 1 n 2 n 3…….. n 4

10. Example (4) Two-stage Procedure A student wanted to go to Dubai (for the Eid holiday) on Friday & come back on Monday. She wants to fly only on Qatar Airways. There are 5 Qatar Airways flights to Dubai on Friday, and 4 Qatar Airways flights back to Doha on Monday. She also have the opportunity to hitch a ride with a friend to Dubai and back. In how many ways she can complete her round trip? Solution: The number of ways to accomplish the first task of going to Dubai = 5 + 1 = 6 The number of ways to accomplish the second task of coming back to Doha = 4 + 1 = 5 The number of ways to complete the round trip = 6(5) = 20

11. Example (5) A man wants to buy school bags for his 5 children, he found out, when he visited the shop that the bags come in three sizes, small, mediums and large, and in two colors; black and brown. He wants to take home samples to show his children, before buying. How many samples should he take to cover all types? Solution: Study the chart called tree diagram on the next slide. The number of the types of bags with regard to size = 3 The number of the types of bags with regard to color = 2 The number of samples = the number of the types of bags with regard to size & color = 3(2) = 6

12. Tree Diagram Small Large Medium Brown black Brown First Stage: Selecting size Second Stage: Selecting color

13. Example (6) A three stage procedure How many three-letters words, meaningful or otherwise, can be made from the letters A,B,C and D, with no letter is repeated in any word? Solution: Study the chart called on the next slide. The number of positions in the word that each letter can assume is three. For the first position, we can select any letter from the four, say the letter “A”. Doing that, we can fill the second position with any of the remaining three letters “B, C or D”. Doing so, say selecting B, the third position can be filled with any of the remaining two letters “C or D”. Thus there are four ways in selecting the first position (completing the first stage), three ways in selecting the second position (completing the second stage), and two ways in selecting the third position (completing the third stage). So, we have three-stage procedure completed correspondingly, in 4, 3 and 2 ways. Thus the number of ways to complete the whole procedure is equal to: 4(3)(2) = 24. This is the number of words that can be formed from the four letters with no letter repeated in any word.

14. D B First Position 4 ways A B C D Second Position 3 ways C C D A Third Position: 2 ways complete the tree diagram for this stage B AD C B A C D

15. Question (1) 4-stage procedures a. Three highways connect towns A & B, two connects B & C, 6 connects C & D and 5 connects D & E. How many distinct routes one can choose from to drive from A to E ? Answer 180 routes b. Two highways connect towns A & B, two connect B & C, 2 connects C & D, 2 connects D & E, 2 connect E& F 2 connect F& G and 2,. How many distinct routes one can choose from to drive from A to G ? Answer 2 6 = 64 routes

16. Question (2) a. A quiz consists of 4 multiple-choice questions with three choices for each. In how many ways it can be answered ? Answer : we have 4 stages procedure, with each stage doable in three ways. Answer: 3(3)(3)(3)=3 4 = 81 b. A quiz consists of 6 true-false questions. In how many ways it can be answered ? Answer : we have 6 stages procedure, with each stage doable in two ways. Answer: 2 6 = 64 c. A quiz consists of 4 multiple-choice questions with three choices for each and 6 true-false questions. In how many ways it can be answered ? Answer : we have 2 stage-procedure, the first is the answering the multiple- choice questions (which is doable in 81 ways) and the second answering the true-false questions (doable in 64 ways). Continue?

17. Question (3) a. From 6 distinct positive integers, how many 4-digit numbers, we can make, provided no integer appears twice in the number? b. From 6 different letters, how many 4-letter words (meaningful or otherwise), we can make provided no letter appears twice in the word? Solution: In both questions, there are 4 places to be filled by three distinct objects out of 6. There are 6 objects to choose from to fill the first place ( 6 ways to perform the first stage ). Once the first place is filled, there will be 5 objects to choose from to fill the second place ( 5 ways to perform the second stage ). Once the second place is filled, there will be 4 objects to choose from to fill the third place to ( 4 ways to perform the third stage ). Once the third place is filled, there will be 3 objects to choose from to fill the fourth place ( 3 ways to perform the fourth stage ). Thus, we can make 6(5)(4)(3) = 360 four-digit numbers ( four-letter words)

18. Question (4) From 4 distinct positive integers, how many 4-digit numbers, we can make, provided no integer appears twice in the number? Solution: In both questions, there are 4 places to be filled by 4 distinct objects. There are 4 objects to choose from to fill the first place ( 4 ways to perform the first stage ). Once the first place is filled, there will be 3 objects to choose from to fill the second place to ( 3 ways to perform the second stage ). Once the second place is filled, there will be 2 objects to choose from to fill the third place to ( 2 ways to perform the third stage ). Once the third place is filled, there will be only one object left to fill the fourth place to ( 1 way to perform the fourth stage ). Thus, we can make 4(3)(2)(1) = 24 four-digit numbers ( four-letter words)

19. Notation n! =1(2)(3)(4)……(n) = n(n-1)(n-2)….(3)(2)(1) 0! = 1 Notice that: (n + 1)! = (n+1)n(n-1)(n-2)…(3)(2)(1) = (n+1) n! = (n+1) n (n-1)! Examples: 5! = 5(4)(3)(2)(1) = 120 5! =5(4!) = 5(4)(3!)=5(4(3)(2!)

20. Permutations of n distinct objects I. The number of permutations of n distinct objects taken r at a time p(n,r) = n! / (n - r)! Result: p(n,n) = n! / (n - n)! = n! / 0! = n! / 1 = n! illustration: 1. p(6,4)! = 6! / (6-4)! = 6! / 2! = 6(5)! / 2! = 6(120) / 2 = 3(120) = 360 2. p(5,5) = 5! = 120

21. Example (7) Let A = { a,b,c,d,e,f} Compute the number of permutations of the set A taken 2 at a time Solution: We have: n = 6 and r = 2 P(6,2) = 6! / (6-2)! = 6! / 4! = 6(5)4! / 4! = 30

22. Question (5) a. From 6 distinct positive integers, how many 4-digit numbers, we can make, provided no integer appears twice in the number? b. From 6 different letters, how many 6-letter words (meaningful or otherwise), we can make provided no letter appears twice in the word? Solution: a. we are arranging 6 objects taken 4 at a time. The number of such arrangement = p(6,4)= 6! / (6-4)! = 6(5)(4)(3)2!/2! = 360 b. we are arranging 6 objects taken 6 at a time. The number of such arrangement = p(6,6) = 6! = 6(5)(4)(3)(2)(1)=720

23. Question (6) There are 25 students in this class. Out of them 16 students has done very well in the first exam. How many ways Dr. Fuad can choose a 5 team leaders of projects from these 16 students if there are five different projects ? Solution: The number of ways of selecting 5 students out of 16 students to lead 5 different projects = p(16,5) = 16! / (16 – 5 )! = 16! / 11! = 16(15)(14)(13)(12) =524160

24. Permutations of n not all distinct objects Taken n at a time I. The number of permutations of n objects of which n 1 are alike and of one kind, n 2 are alike and of another kind, n 3 are alike and of third kind, …….and n m are alike and of yet another kind, such that: N = n 1 + n 2 + n 3 + ………….+ n m is: (n)! / (n 1 )! (n 2 )! (n 3 )! …….. n m !

25. Example (8) Compute the number of permutations that can be formed from all the letters of the name: Saudi Arabia Solution: We have: n = 11 ( the number of objects (letters) involved) n 1 = 4 ( the number of A’s: 4 alike objects of one kind) n 2 = 2 ( the number of I’s: 4 alike objects of one kind) n 3 = 1 (the number of S’s), n 4 = 1 (the number of u’s) n 5 = 1 (the number of d’s), n 6 = 1 (the number of r’s) n 7 = 1 (the number of b’s) Thus the number of permutation = (11)! / 4! 2! 1! 1! 1! 1! 1! =11(10)9(8)(7)(6)(5)(4!)/4! 2= 11(10)(9)(8)(7)(3)(5)=831600

26. Combination of n distinct objects I. The number of combinations of n distinct objects taken r at a time C(n,r) = n! / r! (n - r)!, where r ≤ n Result: C(n,n) = n! / n! (n - n)! = n! / n! 0! = 1 Examples: 1. C(6,4)! = 6! / 4! (6-4)! = 6! / 4! 2! = 6(5) 4! / 4! 2! = 15) / 2 2. C(5,5) = 1

27. Important Note The Difference between Permutation & Combination I. Permutation of the element of the set {a,b, c, d} taken 3 at a time: We can pick the three elements a, b and c and arranged them in different orders, with each ordr counted as a different choice. The choices abc, acb, bca, bac, cab, cba are counted as different ways (They are different permutations: Just like making words of the 4 distinct given letters) p(4,3) = 4! / (4 - 3)! = 4! = 4(3)(2)(1) = 24 Checking with our eyes!!: The permutations are: abc, acb, bca, bac, cab, cba dbc,dcb, bcd, bdc, cdb, cbd, adc, acd, dca, dac, cad, cda abd, adb, bda, bad, dab, dba I. Combination of the element of the set {a,b,c, d} taken 3 at a time: The choices {a,b,c}, {a,c,b}, {b,c,a}, {b,a,c}, {c,a,b}, {c,b,a}, are counted as one way (They are the same combinations: Just like counting the number of 3- member subset of the set {a,b,c, d} ) and C(4,3) = 4! / 3!(4 - 3)! = 4 Checking with our eyes!!: The combinations are: {a,b,c}, {a,b,d}, {c,b,d}, {c,a,d},

28. Example (9) From a number of 8 people, 4 people are to be selected for some task. Compute the number of ways to do that Solution: We have: n = 8 r = 4 The number of ways to select 6 people out of 8 = C (8,4) = 8! / 4! (8-4)! = 8(7)(6)(5)4! / 4! (4)(3)(2)(1) =2(7)(5) = 70

29. Question (7) There are 25 students in this class. Out of them 16 students has done very well in the first exam. How many ways Dr. Fuad can choose a 5 team leaders committee ? Solution: The number of ways of selecting 5 students out of 16 students = C(16,5) = 16! / 5! (16 – 5 )! = 16! / 5! 11! = 16(15)(14)(13)(12) / 5(4)(3)(2)= 2(14)(13)(12) = 4368