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2. 2 Warm Up 1.List all combinations of Roshanda, Shelli, Toshi, and Hector, taken three at a time 2. In how many ways can 5 people be seated in a row of 5 chairs?

3. Copyright © Ed2Net Learning Inc.3 4. In how many ways can 3 representatives be chosen from a group of 11 people? 5. In how many ways can a 5player team be chosen from 16 people? 3. In how many ways can three clarinet players be seated in first, second, and third seats in the orchestra? Warm Up

4. Copyright © Ed2Net Learning Inc.4 Let’s summarize what we have learnt in the previous lesson 1. The fundamental counting principle says that, the the total number of different ways a task can occur is: n1 * n2 * n3 * ………………… * nr where n1, n2, n3..nr are the tasks performed at first, second and rth stages 2. Factorial is a shorthand notation for a multiplication process. The Fundamental Counting Principle says that the number of ways to choose n items is n ! n! = n * (n-1) * (n-2) *……………*3 * 2 * 1 3. Any factorial less than n! is a factor of n!

5. Copyright © Ed2Net Learning Inc.5 5. The number of ways (or combinations) in which r objects can be selected from a set of n objects, where repetition is not allowed, is denoted by, or C(n, r) n C r n! n C r = (n-r)! r! 6. Pascal’s Triangle illustrates the symmetric nature of a combination. i,e C(n,r) = C(n,n-r) Recap 4. A combination focuses on the selection of objects without regard to the order in which they are selected

6. Copyright © Ed2Net Learning Inc.6 10. The sum of all combinations : C(n,0) + C(n, 1) +... + C(n, n) = 2^n 11. Relationship between permutation and combination is p n r n C r = r! Recap 7. C(n, 1) = n 8. C(n, 0) = 1 9. C(n, n) = 1

7. Copyright © Ed2Net Learning Inc.7 Concept of Permutation: The letters a, b and c can be arranged in six different ways abc acb bac bca cab cba Each of these arrangements is called the permutations of the letters a, b and c

8. Copyright © Ed2Net Learning Inc.8 The symbol n P r is used to indicate the number of permutations n objects taken r at a time. n P r = n (n -1)(n -2)…….[n-(r-1)]=n!/ (n - r)! When r = n, last factor of the in the above equation n-(r-1) = n-(n-1) = 1 n P r = n (n -1)(n -2)…….1= n! r factors

9. Copyright © Ed2Net Learning Inc.9 A permutation focuses on the arrangement of objects with regard to the order in which they are arranged. A combination focuses on the selection of objects without regard to the order in which they are selected. The distinction between a combination and a permutation has to do with the sequence or order in which objects appear. Permutation and Combination - Difference

10. Copyright © Ed2Net Learning Inc.10 Since the order does not matter in combinations, there are clearly fewer combinations than permutations. The combinations are contained among the permutations. They are a subset of the permutations Relation between permutation and combination

11. Copyright © Ed2Net Learning Inc.11 Example Find n P r for n = 7 and r = 5. n P r = 7 P 5 = = = 7.6.5.4.3.2.1/2.1 = 7.6.5.4 = 720 nPr = n!/(n-r)! 7! (7 – 5)! 7! 2! n! (n – r)!

12. Copyright © Ed2Net Learning Inc.12 Example Find n P r for n = 4 and r = 4. n P n = n! 4 P 4 = 4! = 4.3.2.1 = 24 nPn = n!

13. Copyright © Ed2Net Learning Inc.13 Example Find the number of permutations of the three letters b, c and d. Solution: The number of permutations of n objects is n! n P n = n! 3 P 3 = 3! = 3.2.1 = 12 The number of Permutations of n objects is n!

14. Copyright © Ed2Net Learning Inc.14 Example From a set of 8 different books, 5 are to selected and arranged on a self. How many arrangements are possible? Solution: Find the number of permutations of 8 books taken 3 at a time: n P r = 8 P 5 = 8!/(8-5)! = 8.7.6.5.4.3.2.1/3.2.1 = 8.7.6.5.4 = 6720 The number of permutations of n objects taken r at a time is nPr. n! (n – r)!

15. Copyright © Ed2Net Learning Inc.15 If a set of n elements has n 1 elements of one kind, n 2 of another kind alike and so on, then the number of permutations, P, of the n elements taken n at a time is given by: P = n! n 1 !.n 2 !.......

16. Copyright © Ed2Net Learning Inc.16 Example Find the number of ways the letters in the word BEEKEEPER can be arranged? Solution: There are 9 letters, 5 of which are E’s, 1 is B, 1 is K and 1 is P. P = P = 9!/5!.1!.1! = 9.8.7.6.5.4.3.2.1/5.4.3.2.1.1.1 = 9.8.7.6 = 3024 P = n! n 1 !.n 2 !....... n! n 1 !.n 2 !...

17. Copyright © Ed2Net Learning Inc.17 Your Turn Evaluate: 1.5! 2.10!/8! 3.6!/(6-2)! 4. 5 P 5 5. 9 P 3

18. Copyright © Ed2Net Learning Inc.18 Your Turn Tell the number of ways the letters in each word can be arranged. 6.TOP 7.NOON 8.SUPPER 9.In how many ways 7 people can be lined up in a row? 10. In how many ways 6 different books can be arranged on a self?

19. Copyright © Ed2Net Learning Inc.19 Refreshment Time

20. Copyright © Ed2Net Learning Inc.20 Lets play a Game www.miniclip.com/games/monkey-lander/en/

21. Copyright © Ed2Net Learning Inc.21 1. In how many ways can 3 cards from a deck of cards be laid in a row face up?

22. Copyright © Ed2Net Learning Inc.22 2. a) How many vowels are there? How many consonants are there? b) In how many ways can you choose at random a vowel and a consonant?

23. Copyright © Ed2Net Learning Inc.23 3. Show that 6 P 4 = 6( 5 P 3 )

24. Copyright © Ed2Net Learning Inc.24 Let Us Review The symbol n P r is used to indicate the number of permutations n objects taken r at a time. n P r = n!/ (n - r)! The numbers of n objects is n!. If a set of n elements has n 1 elements of one kind, n 2 of another kind alike and so on, then the number of permutations, P, of the n elements taken n at a time is given by: P = n! n 1 !.n 2 !.......

25. Copyright © Ed2Net Learning Inc.25 You did Great !!!! Keep up the good work