4-2 Factorials and Permutations
Imagine 3 animals running a race: How many different finish orders could there be? D H S FINISHFINISH
1 st D 2 nd 3 rd Order H S H S D S H D S H S D D H DHS DSH HDS HSD SHD SDH
THEREFORE: There are 6 possible permutations (ordered lists) for the race. This technique will be too cumbersome for questions with any complexity…… So….
Another way: 1st2nd3rd How many choices are there for first place? 3 3 Second place? 2 2 Third place? X 2 X 1 = 6 (This can be compressed even further)
Note: 3 X 2 X 1 can be compressed into Factorial Notation: 3! n! = n X (n – 1) X (n – 2) X … X 3 X 2 X 1 Ex: 5! = 5 X 4 X 3 X 2 X 1 = 120
Simplify (on board) 8! = 10! = 7!
The senior choir has a concert coming up where they will perform 5 songs. In how many different orders can they sing the songs?
In how many ways could 10 questions on a test be arranged if a) there are no limitations b) the Easiest question and the most Difficult question are side by side c) E and D are never side by side
a) No limitations XXXXXX XXX = 10! b) E and D are side by side XXXXXX XXX E D = 9! X 2 c) E and D are never side by side 10! – 9! X 2
Permutation (when order matters) A permutation is an ordered arrangement of objects (r) selected from a set (n).
P(n,r) (also written as n P r ) represents the number of permutations possible in which r objects from a set of n different objects are arranged. With the 3 animal race, it would have been 3 objects (n = 3), permute 3 objects (r = 3) P(3,3) or 3 P 3
How many first, second, and third place finishers can there be with 5 animals? 5 1st2nd3rd 43 5 X 4 X 3 = 60 (way too many to tree) P(5,3) or 5 P 3
We want to use the factorial notation…. 5 animals, 3 spots… 5 X 4 X 3 X 2 X 1 2 X = 5 X 4 X 3 = 60 5! 2! = 5! (5 – 3)! = n! (n – r)! P(n,r)
How many different sequences of 13 cards can be drawn from a deck of 52? 52 P 13 = 52 P 13 =
Pg 239 [1-4] odd 7,9,10,11 14,15,19,20