MATH- PERMUTATION (Circular, Distinguishable, etc).pptx

Finding the number of permutation in an object

1. Permutation

2. A small padlocked treasure chest was found in an abandoned island. In order to open this chest containing precious jewels, a 4- letter password without repetition must be unlocked using the letters A, B, C, and D. If you are going to list down all the possible codes, how many codes will there be?

3. Permutation Permutation is an arrangement of objects in a specific order. Permutation also refers to any one of all possible arrangements of the elements of the given set. Permutation is when the order or arrangement is IMPORTANT .

4. Permutation WITHOUT restrictions

5. Example #1 Find the number of ways to arrange the letters ABC.

6. By listing method there are 6 ways. A-B-C A-C-B B-A-C B-C-A C-A-B C-B-A

7. !!! Instead of listing all the possible arrangement there is an easy way to know how many possible arrangements are there.

8. Factorial (!) Means to multiply a series of descending natural numbers. It's a shorthand way of writing numbers the product of all positive integers less than or equal to n n n!     1 1 1 1 2 2 ×  1 = 2 ×  1! = 2 3 3 ×  2 × 1 = 3 ×  2! = 6 4 4 ×  3 × 2 × 1 = 4 ×  3! = 24 5 5 ×  4 × 3 × 2 × 1 = 5 ×  4! = 120

9. Solution: Given: n= 3 n! = n(n-1)(n-2)….(n-r+1) 3! = 3 (3-1)(3-2) 3! = (3)(2)(1) 3! = 6 Therefore , letters ABC can arrange in 6 ways. .

10. Try this! 5! 2! 8! 5x4x3x2x1 2x1 8x7x6x5x4x3x2x1 = 120 = 2 = 40, 320

11. The number of permutations of n things taken n at a time is given by: nPn = = = n !   CASE # 1 :

12. Example #2: Maia has 6 potted plants. In how many ways can she arrange these plants in a row?

13. Solution: Given: n= 6 books nPn = 6! 6 P 6 = (6)(5)(4)(3)(2)(1) 6 P 6 = 720 ways

14. The number of permutations of n things taken r at a time is given by: nPr= !   CASE # 2 : n= no. of objects r= no, of position

15. Example # 3: How many four-digit numbers can be formed from the numbers 1, 3, 4, 6, 8, and 9 if repetition of digits is not allowed?

16. Solution: There are four position to be filled up. How many choices are there for the first blank? _____ _ ______ x ____________ x ____________ x _____________

17. Solution: There are four position to be filled up. _____ 6 ______ x ____________ x ____________ x _____________ How many choices are there for the second blank?

18. Solution: There are four position to be filled up. _____ 6 ______ x _____ 5 _____ x ____________ x _____________ How many choices are there for the third blank?

19. Solution: There are four position to be filled up. _____ 6 ______ x _____ 5 _____ x _____ 4 ____ x _____________ How many choices are there for the fourth blank?

20. Solution: There are four position to be filled up. _____ 6 ______ x _____ 5 _____ x _____ 4 ____ x _____ 3 ______ = 360 four- digit numbers will be formed

21. Solution: = = = = = 360  

22. Example # 4: Fifteen cars enter a race. In how many ways can the trophies for first, second and third place be awarded?

23. Permutation with like Objects (Distinguishable Permutation) is denoted by: P= !   CASE # 3 : n= no. of objects n k = represent the letters or object that is repeated.

24. Example #5: How many ways can the letters of the word “TAGAYTAY” be arranged?

25. Solution: Given: TAGAYTAY Letters in all= 8 T’s= 2 A’s= 3 Y’s= 2 P= P= P= P= 1680  

26. Example #: 6 In how many ways can the letters in the word “TALLAHASSEE” be arranged?

27. Solution: TALLAHASSEE Given: “TALLAHASSEE” Letters in all= 11 A’s= 3 L’s= 2 S’s= 2 E’s = 2 P= P= P= P= 831, 600  

28. Circular permutation - is the total number of ways in which n distinct objects can be arranged around a fixed circle. The number of ways to arrange n distinct objects along a fixed circle is P= (n-1)! ! CASE # 4 :

29. Example # 7: In how many ways can 4 people be seated around a circular table?

30. Example # 8: In how many ways can 5 boys and 4 girls be seated around a circle table without any restriction?

31. Permutation WITH restrictions

32. Example # 8: In how many ways can 3 boys and 4 girls be seated in a table; Without restriction The parents stand together All the females stand together

33. Solution: there are no restrictions n= 7! n= 7 x 6 x 5 x 4 x 3 x 2 x 1 n= 5040 Therefore, the family members can be line up in 5040 ways.

34. Solution: the parents stand together n= 6! 2! n= (6x 5 x 4 x 3 x 2 x 1) (2 x 1) n= (720)(2) n= 1440 Therefore, the family members can be line up in 1440 ways if the parents stand together.

35. Solution: all the females stand together n= 4! 4! n= (4 x 3 x 2 x 1)(4 x3 x 2 x 1) n= (24)(24) n= 576 Therefore, there are 576 different ways the family can line up if the females stand together.

36. Example # 9: How many three-digit numbers greater than 300 may be formed using the numbers 1, 2, 3, 7 and 9 if repetition of digit is allowed?

37. Solution:

38. THANK YOU! 