Factorials and Permutations. In how many ways can a senior choir rehears five songs for an upcoming assembly. There are a total of 5 songs to choose from.

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Presentation transcript:

Factorials and Permutations

In how many ways can a senior choir rehears five songs for an upcoming assembly. There are a total of 5 songs to choose from first To make the second choice, there are 4 songs left. To make the third choice, there are 3 songs left To make the fourth choice, there are 2 songs left. To make the fifth choice, there is 1 song left. These are not mutually exclusive choices, therefore we can multiply them together. 5x4x3x2x1 = 120

This is called a factorial N! = n x (n – 1) x … x 3 x 2 x 1 Read as n factorial Try 4!, 8! 10! 83 ! 5! 79 ! You can cancel out some of the common values so that it fits in your calculator.

Indirect method How many ways could 10 questions on a test be arranged, if the easiest and most difficult question are a) side by side b) not side by side a) If the easiest and hardest questions are a unit, there are nine items to be arranged. The two questions can be arranged in 2 ways (2!). 9! X 2! = a) If we use the indirect method. The number of arrangements with the easiest and most difficult questions separated is just equal to the total number of possible arrangements minus the number with the two questions side by side (what we did last question)

A permutation is an arrangement of all the items in a definite order. The total number of permutations is denoted by n P n or P(n,n). There are n ways of choosing the first term, n-1 ways of choosing the second term. n P n = n x n-1 x … x 3 x 2 x 1 Find 8 P 8 = 8! = 40320

How many ways are there to choose a President and a vice president from this entire class? 30 x 29 ways to choose a pres and v.p

A permutation of n distinct items taken ‘r’ at a time is an arrangements of r of the n items in a definite order. These permutations are called r-arrangements of n items. The total number of possible arrangements of n items. It is denoted by n P r or P(n,r) N ways of choosing the first, n –1 ways of choosing the second, n – r + 1 ways of choosing the last. N(n-1)(n-2)…(n-r+1).

More convenient to write This is the number of permutations of r items taken from n distinct items

Example. In a card game, each player is dealt 13 cards face down that can be turned up and used one by one during the game. How many different sequences or cards could a player have? You need to take 13 cards from a deck of P 13 = 52! / (52 – 13)! = x Homework pg 239 # 1,2ace,4ad, 6b, 8, 10, 14, 15