Chapter 5 Section 5 Permutations and Combinations Basic

Chapter 5 Section 5 Permutations and Combinations

Basic Idea 1. Counting problems in which the only restriction is that there are no repetitions allowed (with the exception of tossing a coin). (This implies that once you selected an element/object, you can not select it again. This is often referred to as selecting without replacement) 2. Then the big question is: Does the order of the elements in the set matter?

Does the order of the elements in the set matter? • If yes, then use the Permutation • If no, then use the Combination • Remember: No Repetitions (with the exception of tossing a coin).

Permutation Notation P(n , r) = n • (n – 1) • (n – 2) • (n – 3) • … • (n – r +1) n = the number of objects that you can choose from. r = the number of objects that you select (without replacement) Note: 1. 2. 3. n is a positive integer ( 1, 2, 3, 4, … ) r is a whole number ( 0, 1, 2, 3, 4, … ) n>r

Steps to find a permutation by hand Steps: 1. Create ‘r’ underlines. 2. Put the value of ‘n’ in the first underline. 3. Rule: The next underline always contains the value that is one less than the value in the previous underline. Repeat step 3 until all underlines are filled with values 4. 5. Multiply all the values together (i. e. the multiplication principle)

Permutation Example • Recall Exercise 15 from page 228 – – 30 member football team Must select a captain and assistant captain Select elements (names) without replacement Order of the elements in the set does matter since person selected first is captain – Solution: 30 • 29 = 870 • This is a permutation problem since (1) we selected elements without replacement and (2) the order of the elements in the set mattered! – Rewritten Solution: P( 30 , 2 ) = 30 • 29 = 870

Two Special Permutations 1. Definition: For any value n, P( n , 0 ) = 1 2. Factorials Notation: Definition n! n! = n • (n – 1) • (n – 2) • … • 3 • 2 • 1 When n = r, then P(n , r) = n! Note: Definition: 0! = 1

Factorial Example • An owner of 5 stores wants to create a ranked list of their stores based on the yearly sale. How many possible lists could be made? 5 ! = 5 • 4 • 3 • 2 • 1 = 120 which is equivalent to: P( 5 , 5 ) = 5 • 4 • 3 • 2 • 1 = 120

Combination Notation C(n , r) = P(n , r) / r! n = the number of objects that you can choose from. r = the number of objects that you select (without replacement) Similar rules apply to the possible values of n and r

Why is C(n, r) = P(n, r) / r! • It is best to view it as P(n, r) = C(n, r)·r! • Take the following Example: Take the letters A, B, C, and D and pick three letters. So n = 4 and r = 3

Pick Three Letter Example (All possible permutation) ABC ACB BAC BCA CAB CBA ABD ADB BAD BDA DAB DBA ACD ADC CAD CDA DAC DCA BCD BDC CBD CDB DBC DCB

Pick Three Letter Example (All different combinations in box) ABC ACB BAC BCA CAB CBA ABD ADB BAD BDA DAB DBA ACD ADC CAD CDA DAC DCA BCD BDC CBD CDB DBC DCB

Pick Three Letter Example (Permutations within one of the combinations) ABC ACB BAC BCA CAB CBA ABD ADB BAD BDA DAB DBA ACD ADC CAD CDA DAC DCA BCD BDC CBD CDB DBC DCB

Pick Three Letter Example (Number of rows times number of columns equals the number of possible permutations) ABC ACB BAC BCA CAB CBA ABD ADB BAD BDA DAB DBA ACD ADC CAD CDA DAC DCA BCD BDC CBD CDB DBC DCB

Pick Three Letter Example • There are 24 total words that you could create when the order of the letters are important. P(4, 3) = 24 • There were 4 combinations possible (listed in the vertical box) when order of the letters did not matter. C(4, 3) = 4 • For each combination, there were six possible ways to arrange three letters (listed in the horizontal box). 3· 2· 1 = 3!. • Thus P(4, 3) = C(4, 3) · 3! • Now solve for C(4, 3). • C(4, 3) = P(4, 3) / 3!

Example • How many ways can we select 2 books from a set of 9 books? – No repetitions – Order of the elements in the set doesn’t matter C( 9 , 2 ) = ( 9 • 8 ) / ( 2 • 1 ) = 36

Useful Pattern to Combinations For: C( 5 , r ) r C( 5 , r ) 0 1 1 5 2 10 3 10 4 5 5 1 For: C( 6 , r ) r C( 6 , r ) 0 1 2 3 4 5 6 15 20 15 6 1

Combinations are useful for: • Poker Hand Problems • Tossing a Coin (Section 6) • Drawing Balls Out of an Urn (Section 6)

Recall: A Standard Deck of 52 Cards There are: • 4 Suits : (Clubs , Diamonds, Hearts, Spades) • 13 Denominations: (2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A) • 2 Colors : – ( Red : Diamonds & Hearts , Black : Clubs & Spades) • So – 13 cards of each suit – 4 cards of each denomination (one in each suit) – 26 red cards ( Diamonds & Hearts) – 26 black cards ( Clubs & Spades )

Poker Hands • Poker hand is 5 cards selected from a deck of 52 cards. – Drawing without replacement (No repetitions) – Order of the elements in the set (i. e. the cards in your hand) doesn’t matter! • Thus we can use combinations to count the number of hands in a poker hand! • Hints: (1) (2) r=5 Trick is to determine what n is, so think of the restrictions placed on n

Problem: • How many different poker hands consists of the 5 cards being either a 2 , or 7? Solution: • There are 8 cards to choose from (four 2’s and four 7’s) • We choose 5 cards C( 8 , 5 ) = 56