0.5 – permutations & combinations

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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. - PowerPoint PPT Presentation

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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.

Permutation Formula – The number of permutations of n objects taken r at a time is the quotient of n! and (n – r)!

• 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)!

• 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.

• 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!

• 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!

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

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)!

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)!

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!

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 ∙ ∙ ∙ ∙ ∙ ∙

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 ∙ ∙ ∙ ∙ ∙ ∙

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 ∙ ∙

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∙ ∙

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?

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!

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!

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 ∙ ∙ ∙ ∙ ∙ ∙ ∙

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 ∙ ∙ ∙ ∙ ∙ ∙ ∙

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 ∙ ∙ ∙ ∙ ∙ ∙ ∙

Permutations with RepetitionThe number of permutations of n objects of which p are alike and q are alike is

n!_ p!q!

Permutations with RepetitionThe number of permutations of n objects of which p are alike and q are alike is

n!_ p!q!

Ex. 3 How many different ways can the letters in the word MISSISSIPPI be arranged?

Permutations with RepetitionThe number of permutations of n objects of which p are alike and q are 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.

Permutations with RepetitionThe number of permutations of n objects of which p are alike and q are 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!

Permutations with RepetitionThe number of permutations of n objects of which p are alike and q are 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!

Permutations with RepetitionThe number of permutations of n objects of which p are alike and q are 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 ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙

Permutations with RepetitionThe number of permutations of n objects of which p are alike and q are 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 ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙

3

2

5

11 5 3 7 5 4 3∙ ∙ ∙ ∙ ∙ ∙ 2

11 5 3 7 5 4 3∙ ∙ ∙ ∙ ∙ ∙ = 34,650 2

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