# Permutations - Arrangements Working out the number of elements in a sample space without having to list them IntroductionIntro ExerciseFactorials Factorial.

## Presentation on theme: "Permutations - Arrangements Working out the number of elements in a sample space without having to list them IntroductionIntro ExerciseFactorials Factorial."— Presentation transcript:

Permutations - Arrangements Working out the number of elements in a sample space without having to list them IntroductionIntro ExerciseFactorials Factorial Practice

Counting multiple events (including choices from different groups) The number of different possibilities from a multiple event situation can be determined by: (a) Listing all the possibilities (using common sense, a table or a tree diagram) (b) Using a box diagram.

Counting Arrangements (Permutations) With The Box Method Notes: Instead of notes to copy into your book, we are going to work through 7 questions and then you can ask for extra notes if you feel that you need them. Remember to stick the example sheet into your workbook.

Counting multiple events (including choices from different groups) Example 1 How many different possibilities are possible when tossing 3 coins HHHHHT HTHHTT THHTHT TTHTTT Method 1: Counting by listing Method 2: Counting using the box method 8 possibilities Coin 1Coin 2Coin 3 222 = 8

Counting multiple events (including choices from different groups) Example 2 How many different possibilities are possible when rolling 3 dice. Method 1: Counting by listing Method 2: Counting using the box method D 1D 2D 3 666 = 216 Not plausible since too many alternatives

Counting multiple events (including choices from different groups) Example 3 How many different possibilities are possible when choosing 3 courses off a menu with 3 different entrees, 7 main courses and 4 desserts. EMD 374 = 84

Counting multiple events (including choices from different groups) Example 4 How many different standard number plates are possible in Victoria where we have 3 letters and 3 numbers. L1L2L3N1N2N3 26 10 = 17,576,000 L1L2L3N1N2N3 2526 10 = 16,900,000 Extra Question – How many standard number plates are available to the public?

Counting Arrangements (Arrangement = Ordered group OR Permutation) Example 5 – Arranging Books (a) How many different arrangements (ordered groups or permutations) are possible of 6 books on a shelf B1B2B3B4B5B6 654321 = 720

Counting Arrangements (Arrangement = Ordered group OR Permutation) Example 5 – Arranging Books (b) If 6 books are chosen from 10 titles and placed on a shelf, how many different arrangements (ordered groups or permutations) are possible on the shelf. B1B2B3B4B5B6 1098765 = 151,200

Counting Arrangements (Arrangement = Ordered group OR Permutation) Example 6 – Arranging Letters How many ways can the letters in the word CAPTION be arranged: (a) With no conditions on letter placements? (b) Starting with a vowel? (c) Starting and ending with a consonant? L1L2L3L4L5L6L7 7654321 = 5040 VL1L2L3L4L5L6 3654321 = 2160 C1L1L2L3L4L5C2 4543213 = 1440

Counting Arrangements (Arrangement = Ordered group OR Permutation) Example 6 – Arranging Letters How many ways can the letters in the word CAPTION be arranged: (d) With vowels and consonants in alternate positions? (e) Not starting with C? (f) Starting with the C and ending with N? C1V1C2V2C3V3C4 4332211 = 144 Not CL1L2L3L4L5L6 6654321 = 4320 CLL2L3L4L5N 1543211 = 120

Counting Arrangements (Arrangement = Ordered group OR Permutation) Example 6 – Arranging Letters How many ways can the letters in the word CAPTION be arranged: (g) With consonants together? C1C2C3C4V1V2V3 4321321 = 144 V1C1C2C3C4V2V3 3432121 = 144 V1V2C1C2C3C4V3 3243211 = 144 V1V2V1C1C2C3C4 3214321 = 144 Total number of arrangements = 144 + 144 + 144+ 144 =576

Counting Arrangements (Arrangement = Ordered group OR Permutation) Example 7 – Flag Problem A flag is made up of 4 coloured panels. If the possible colours are red, blue, white, green and yellow and no colour is used more than once how many arrangements are possible: (a) With no conditions? (b) With white at the start and any colour except green at the end? (c) With yellow as one of the stripes?

Using the box method for working out the number of Permutations Ex 12B p354 Q3 - 15

Factorials – A shortcut for multiplying boxes 3! = 3 x 2 x 1 = 6 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40,320 5! = 120 10! = 3,628,800 7654321 = 7! = 5040

Factorial Fractions 1211109876

Factorial Problems Important Fact 0! = 1 Ex12B p354 Work out Q1 & 2 with a calculator (keyboard Math/Calc) Work out Q1& 2 without a calculator

Factorial Problems Homework Ex12B p354 Confirm your calculations in questions 3 to 15 using the factorial function on your calculator.

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