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Pearson Unit 6 Topic 15: Probability 15-3: Permutations and Combinations Pearson Texas Geometry ©2016 Holt Geometry Texas ©2007.

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Presentation on theme: "Pearson Unit 6 Topic 15: Probability 15-3: Permutations and Combinations Pearson Texas Geometry ©2016 Holt Geometry Texas ©2007."— Presentation transcript:

1 Pearson Unit 6 Topic 15: Probability 15-3: Permutations and Combinations Pearson Texas Geometry ©2016 Holt Geometry Texas ©2007

2 TEKS Focus: (13)(A) Develop strategies to use permutations and combinations to solve contextual problems. (1)(C) Select tools, including real objects, manipulatives paper and pencil, and technology as appropriate, and techniques, including mental math, estimations, and number sense as appropriate, to solve problems.

3

4 The Fundamental Counting Principle
To find the number of ways that events can happen if they follow each other, you multiply the number of ways that the events can occur. N-FACTORIAL Notation: n! The product of an integer multiplied by each integer counting back to 1. 5! = 5  4  3 2  1 = 120

5 Example 1: Find each value. 7! b) 11! c) 0! = 1
= 7  6  5  4  3 2  1 = 5040 = 11  10  9  8 7  6  5  4  3 2  1 = 39,916,800 = 1

6 Explanation of why 0! = 1 https://zero-factorial.com/whatis.html
What is Zero-Factorial? Simple answer: 0! (read "Zero Factorial") is defined to equal 1. Involved answer(s): There are several proofs that have been offered to support this common definition. Example (1) If n! is defined as the product of all positive integers from 1 to n, then: 1! = 1*1 = 1 2! = 1*2 = 2 3! = 1*2*3 = 6 4! = 1*2*3*4 = n! = 1*2*3*...*(n-2)*(n-1)*n and so on. Logically, n! can also be expressed n*(n-1)! . Therefore, at n=1, using n! = n*(n-1)! 1! = 1*0! which simplifies to 1 = 0!

7 Notice the difference between these two situations and formulas—also see next slide.

8 What is the difference between Permutations and Combinations?
Note: this info at end of your notes due to space.

9 Example 2: 6  4  5 = 120 different lunch specials
A deli offers a lunch special if you choose one from each of the following types of sandwiches, side items, and drink choices. How many different lunch specials are possible? 6  4  5 = 120 different lunch specials

10 Example 3: Suppose that a computer generates passwords that begin with a letter followed by 2 digits, such as B37. The same digit can be used more than once. How many different passwords can the computer generate? 26  10  10 = different passwords

11 Example 4: You download 8 songs onto your phone. If you play the songs using the random shuffle option, how many different ways can the sequence of songs be played? 8! = 8 7  6  5  4  3 2  1 40,320 different ways

12 Example 5: The Math club is electing a president, a vice president and a treasurer. How many different ways can the officers be chosen from the 10 members? This is a permutation because the order does matter. n P r = n ! . (n- r)! 10 P 3 = ! . (10-3)! 10 P 3 = 10 ! . 7! Two ways to work this 3,628,800 = 720. 5040 (10)(9)(8)(7)(6)(5)(4)(3)(2)(1) = (10)(9)(8) = (7)(6)(5)(4)(3)(2)(1)

13 Example 6: Twelve swimmers compete in a race. In how many possible ways can the swimmers finish first, second, and third in the race? This is a permutation because the order does matter. n P r = n ! . (n- r)! 12 P 3 = ! . (12-3)! (12)(11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1) = (12)(11)(10) = (9)(8)(7)(6)(5)(4)(3)(2)(1)

14 Example 7 In AP-English, you are required to read 4 books over the summer from a reading list of 12 books. How many different ways can you choose the books from the reading list? This is a combination because it does not matter what order you read the books in.

15 Example 7 continued n C r = n ! . r!(n- r)! 12 C 4 = 12 ! . 4!(12- 4)!
In AP-English, you are required to read 4 books over the summer from a reading list of 12 books. How many different ways can you choose the books from the reading list? n C r = n ! . r!(n- r)! 12 C 4 = ! . 4!(12- 4)! 12  11  10  9  8  7  6  5  4  3 2  1 (4  3  2  1) (8  7  6  5  4  3 2  1) There are 11880/24 = 495 ways to choose 4 books from a reading list of 12.

16 Example 8: A yogurt shop allows you to choose any 3 out of 10 toppings for their daily special between 2:00 – 4:00 pm. How many different possibilities are there? This is a combination because order does not matter. 10 C 3 = ! 3!(10 - 3)! 10  9  8  7  6  5  4  3  2  1 (3  2  1) (7  6  5  4  3 2  1) 10  9  8 = 120 3  2  1 There are 120 ways to choose your toppings.

17 Example 9: Permutation or Combination?

18 Example 10: Permutation or Combination?

19 Example 11:

20 Example 11: continued 15 C 3 = 15 ! . = 455= 3!(15- 3)!
P (choosing 5, 7, and 9) = # of possible ways to choose 5, 7, 9 total # of ways to choose 3 pool balls There is only 1 way to choose 5, 7, and 9 15 C 3 = ! = 455= 3!(15- 3)! P (choosing 5, 7, and 9) = ≈ ≈ 0.22% 455

21 Example 12: To access the computer network at school, the librarian assigns each student a randomly selected 4-digit code. Each digit in the code is a number from 0 to 9, and no number is repeated within the code. Find the probability that you will be assigned a code with the numbers 1, 2,3, and 4 in any order. P (choosing 1, 2, 3, and 4) = # of possible ways to choose 1, 2, 3 and 4 total # of ways to choose 4 numbers There is only 1 way to choose 1, 2, 3, and 4 10 C 4 = ! = 210= 4!(10- 4)! P (choosing 1, 2, 3, and 4) = ≈ ≈ 0.48% 210


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