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Finite Math

Counting Arrangements: Permutations

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- 1. 2.2Counting Arrangements: Permutations
- 2. Counting Arrangements The idea of counting arrangements is to ﬁnd the number of different ways to arrange a set of things. For example, take the set {1, 2, 3}. There are 6 ways to arrange this set into a 3-digit number without using a number twice: 123 132 213 231 312 321 Each of these arrangement is called a “permutation.”
- 3. Permutation In a permutation, order of the arrangement matters. That means 123 is a different arrangement than 321. More on that in a later section. Also, in a permutation you cannot use the same element twice.
- 4. Permutation Formula n – number of things you’re choosing from r – number you’re choosing (if you are unfamiliar with the “!” notation, check out the factorial tutorial) With our previous example, n = 3 (set has 3 elements, 1, 2, 3), and r = 3 (3-digit number). P(3, 3) = 3!/(3-3)! = 3!/0! = 3 x 2 x 1 = 6
- 5. Slot Method To calculate the number of permutations, the formula is given to you. However, if you recognize a problem to be a permutation problem, the “slot method” is a simpler way. Let’s take our previous example – 3-digit number from {1, 2, 3}. Since it’s a 3-digit number, there are 3 slots:
- 6. The number on digit 1 representsthe number of elements “permitted”on that slot. In this case, there are3.(1, 2, 3 can be the ﬁrst digit)Since there can be no repeats, thereare only two elements permitted ondigit 2:And only one left for digit 3.Simply multiply for the ﬁnal answer: 3 x 2 x 1 =6.
- 7. Slot Method forMultiplication Principle Slot method can be applied for multiplication principle problems as well [1.4]. Lets take the Happy Meal example (3 entrées, 3 drinks, 4 toys): 3 x 3 x 4 = 36.
- 8. Quiz 2.2 #1 How many 3-letter permutations can be formed with the set {A, B, C, D, E}? [Try using both the formula and slots] A. 20 B. 60 C. 120
- 9. Quiz 2.2 #1 How many 3-letter permutations can be formed with the set {A, B, C, D, E}? [Try using both the formula and slots] A. 20 B. 60 C. 120 Answer: B
- 10. The letter permutation problem Lets consider the word “inﬁnite.” How many 8- letter arrangements can be formed from the letters of this word? This is a special problem that requires its own trick. Here’s how it goes: First, treat it as a normal 8-slot permutation: 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40320
- 11. However, that’s too many, because some letters are repeated. So, we divide it by the factorial of number of times a letter is repeated. For example, “i” is repeated 3 times, so you divide by 3!. “n” is repeated twice, so divide by 2!: 3 “i”s 2 “n”s Therefore, the answer is 40320/(6*2) = 3360
- 12. Quiz 2.2 #2
- 13. Quiz 2.2 #2 How many 6 letter arrangements can be formed from the word “effect”? A. 30 B. 60 C. 180
- 14. Quiz 2.2 #2 How many 6 letter arrangements can be formed from the word “effect”? A. 30 B. 60 C. 180
- 15. Quiz 2.2 #2 How many 6 letter arrangements can be formed from the word “effect”? A. 30 B. 60 C. 180 Answer: C
- 16. Summary Definition: Permutation (order matters, no repetition) How to find the number of permutations Formula: Slot method The letter permutation problem
- 17. Features 27 Recorded Lectures Over 116 practice problems with recorded solutions Discussion boards/homework help Visit ﬁnitehelp.com to ﬁnd out more For special offers and additional content...Follow us on twitter @ﬁnitehelp Become a fan on Facebook

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