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4-2 Factorials and Permutations. Imagine 3 animals running a race:. How many different finish orders could there be?. FINISH. D. H. S. H. D. H. D. S. S. 2 nd. 3 rd. Order. 1 st. S. DHS. D. H. DSH. S. HDS. H. HSD. D. D. SHD. S. H. SDH. THEREFORE:.
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Imagine 3 animals running a race: How many different finish orders could there be? FINISH D H S
H D H D S S 2nd 3rd Order 1st S DHS D H DSH S HDS H HSD D D SHD S H SDH
THEREFORE: There are 6 possible permutations (ordered lists) for the race. This technique will be too cumbersome for questions with any complexity…… So….
3 2 1 3 2 1 Another way: 1st 2nd 3rd How many choices are there for first place? 3 X 2 X 1 = 6 (This can be compressed even further) Second place? Third place?
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
10! = 7! Simplify (on board) 8! =
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 limitationsb) the Easiest question and the most Difficult question are side by sidec) E and D are never side by side
X X X X X X X X X X X X X X X X X X 9 8 7 6 5 3 2 1 = 10! 10 4 a) No limitations b) E and D are side by side D = 9! X 2 E 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 nPr) 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 3P3
How many first, second, and third place finishers can there be with 5 animals? 5 4 3 1st 2nd 3rd 5 X 4 X 3 = 60 (way too many to tree) P(5,3) or 5P3
We want to use the factorial notation….5 animals, 3 spots… 5 X 4 X 3 X 2 X 1 1 1 5! 2! 2 X 1 1 1 = 5! = 5 X 4 X 3 (5 – 3)! = 60 P(n,r) = n! (n – r)!
How many different sequences of 13 cards can be drawn from a deck of 52? 52P13 = 52P13=
Pg 239 [1-4] odd 7,9,10,11 14,15,19,20