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0.5 – Permutations & Combinations

0.5 – Permutations & Combinations. Permutation – all possible arrangements of objects in which the order of the objects is taken in to consideration. Permutation – all possible arrangements of objects in which the order of the objects is taken in to consideration.

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0.5 – Permutations & Combinations

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  1. 0.5 – Permutations & Combinations

  2. Permutation – all possible arrangements of objects in which the order of the objects is taken in to consideration.

  3. Permutation – all possible arrangements of objects in which the order of the objects is taken in to consideration. Permutation Formula – The number of permutations of n objects taken r at a time is the quotient of n! and (n – r)!

  4. Permutation – all possible arrangements of objects in which the order of the objects is taken in to consideration. Permutation Formula – The number of permutations of n objects taken r at a time is the quotient of n! and (n – r)! P(n,r) = n! (n – r)!

  5. Permutation – all possible arrangements of objects in which the order of the objects is taken in to consideration. Permutation Formula – The number of permutations of n objects taken r at a time is the quotient of n! and (n – r)! P(n,r) = n! (n – r)! • Combinations – a selection of objects in which order is not considered.

  6. Permutation – all possible arrangements of objects in which the order of the objects is taken in to consideration. Permutation Formula – The number of permutations of n objects taken r at a time is the quotient of n! and (n – r)! P(n,r) = n! (n – r)! • Combinations – a selection of objects in which order is not considered. Combination Formula – The number of combinations of n objects taken r at a time is the quotient of n! and (n – r)!r!

  7. Permutation – all possible arrangements of objects in which the order of the objects is taken in to consideration. Permutation Formula – The number of permutations of n objects taken r at a time is the quotient of n! and (n – r)! P(n,r) = n! (n – r)! • Combinations – a selection of objects in which order is not considered. Combination Formula – The number of combinations of n objects taken r at a time is the quotient of n! and (n – r)!r! C(n,r) = n! (n – r)!r!

  8. Ex. 1 There are 10 finalist in an Olympic competition. How many different ways can gold, silver, & bronze medals be awarded?

  9. Ex. 1 There are 10 finalist in an Olympic competition. How many different ways can gold, silver, & bronze medals be awarded? P(n,r) = n! (n – r)!

  10. Ex. 1 There are 10 finalist in an Olympic competition. How many different ways can gold, silver, & bronze medals be awarded? P(n,r) = n! (n – r)! P(10,3) = 10! (10 – 3)!

  11. Ex. 1 There are 10 finalist in an Olympic competition. How many different ways can gold, silver, & bronze medals be awarded? P(n,r) = n! (n – r)! P(10,3) = 10! (10 – 3)! P(10,3) = 10! 7!

  12. Ex. 1 There are 10 finalist in an Olympic competition. How many different ways can gold, silver, & bronze medals be awarded? P(n,r) = n! (n – r)! P(10,3) = 10! (10 – 3)! P(10,3) = 10! 7! P(10,3) = 10 ∙ 9 ∙ 8 ∙ 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1

  13. Ex. 1 There are 10 finalist in an Olympic competition. How many different ways can gold, silver, & bronze medals be awarded? P(n,r) = n! (n – r)! P(10,3) = 10! (10 – 3)! P(10,3) = 10! 7! P(10,3) = 10 ∙ 9 ∙ 8 ∙ 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1

  14. Ex. 1 There are 10 finalist in an Olympic competition. How many different ways can gold, silver, & bronze medals be awarded? P(n,r) = n! (n – r)! P(10,3) = 10! (10 – 3)! P(10,3) = 10! 7! P(10,3) = 10 ∙ 9 ∙ 8 ∙ 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 P(10,3) = 10 ∙ 9 ∙ 8

  15. Ex. 1 There are 10 finalist in an Olympic competition. How many different ways can gold, silver, & bronze medals be awarded? P(n,r) = n! (n – r)! P(10,3) = 10! (10 – 3)! P(10,3) = 10! 7! P(10,3) = 10 ∙ 9 ∙ 8 ∙ 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 P(10,3) = 10 ∙ 9 ∙ 8 = 720

  16. Ex. 2 Horatio works part-time at a local department store. His manager asked him to choose for display 5 different styles of shirts from the wall of the store that has 8 shirts on it to put in a display. How many ways can he choose the shirts?

  17. Ex. 2 Horatio works part-time at a local department store. His manager asked him to choose for display 5 different styles of shirts from the wall of the store that has 8 shirts on it to put in a display. How many ways can he choose the shirts? C(n,r) = n! (n – r)!r!

  18. Ex. 2 Horatio works part-time at a local department store. His manager asked him to choose for display 5 different styles of shirts from the wall of the store that has 8 shirts on it to put in a display. How many ways can he choose the shirts? C(n,r) = n! (n – r)!r! C(8,5) = 8! (8 – 5)!5!

  19. Ex. 2 Horatio works part-time at a local department store. His manager asked him to choose for display 5 different styles of shirts from the wall of the store that has 8 shirts on it to put in a display. How many ways can he choose the shirts? C(n,r) = n! (n – r)!r! C(8,5) = 8! (8 – 5)!5! C(8,5) = 8 ∙ 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 3 ∙ 2 ∙ 1 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1

  20. Ex. 2 Horatio works part-time at a local department store. His manager asked him to choose for display 5 different styles of shirts from the wall of the store that has 8 shirts on it to put in a display. How many ways can he choose the shirts? C(n,r) = n! (n – r)!r! C(8,5) = 8! (8 – 5)!5! C(8,5) = 8 ∙ 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 3 ∙ 2 ∙ 1 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1

  21. Ex. 2 Horatio works part-time at a local department store. His manager asked him to choose for display 5 different styles of shirts from the wall of the store that has 8 shirts on it to put in a display. How many ways can he choose the shirts? C(n,r) = n! (n – r)!r! C(8,5) = 8! (8 – 5)!5! C(8,5) = 8 ∙ 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 56 3 ∙ 2 ∙ 1 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1

  22. Permutations with Repetition The number of permutations of n objects of which p are alike and qare alike is n!_ p!q!

  23. Permutations with Repetition The number of permutations of n objects of which p are alike and qare alike is n!_ p!q! Ex. 3 How many different ways can the letters in the word MISSISSIPPI be arranged?

  24. Permutations with Repetition The number of permutations of n objects of which p are alike and qare alike is n!_ p!q! Ex. 3 How many different ways can the letters in the word MISSISSIPPI be arranged? 11 total letters, 4 I’s, 4 S’s, and 2 P’s.

  25. Permutations with Repetition The number of permutations of n objects of which p are alike and qare alike is n!_ p!q! Ex. 3 How many different ways can the letters in the word MISSISSIPPI be arranged? 11 total letters, 4 I’s, 4 S’s, and 2 P’s. n!_ p!q!

  26. Permutations with Repetition The number of permutations of n objects of which p are alike and qare alike is n!_ p!q! Ex. 3 How many different ways can the letters in the word MISSISSIPPI be arranged? 11 total letters, 4 I’s, 4 S’s, and 2 P’s. n!_ p!q! 11! _ 4!4!2!

  27. Permutations with Repetition The number of permutations of n objects of which p are alike and qare alike is n!_ p!q! Ex. 3 How many different ways can the letters in the word MISSISSIPPI be arranged? 11 total letters, 4 I’s, 4 S’s, and 2 P’s. n!_ p!q! 11! _ 4!4!2! 11 ∙ 10 ∙ 9 ∙ 8 ∙ 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 4 ∙ 3 ∙ 2 ∙ 1 ∙ 4 ∙ 3 ∙ 2 ∙ 1 ∙ 3 ∙ 2 ∙ 1

  28. Permutations with Repetition The number of permutations of n objects of which p are alike and qare alike is n!_ p!q! Ex. 3 How many different ways can the letters in the word MISSISSIPPI be arranged? 11 total letters, 4 I’s, 4 S’s, and 2 P’s. n!_ p!q! 11! _ 4!4!2! 11 ∙ 10 ∙ 9 ∙ 8 ∙ 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 4 ∙ 3 ∙ 2 ∙ 1 ∙ 4 ∙ 3 ∙ 2 ∙ 1 ∙ 2 ∙ 1 5 3 2

  29. 11 ∙ 5 ∙ 3 ∙ 7 ∙ 5 ∙ 4 ∙ 3 2

  30. 11 ∙ 5 ∙ 3 ∙ 7 ∙ 5 ∙ 4 ∙ 3= 34,650 2

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