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An introduction to permutations

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Presentation on theme: "An introduction to permutations"β€” Presentation transcript:

1 An introduction to permutations
Slideshow 60, Mathematics Mr Richard Sasaki, Room 307

2 Objectives Understand the meaning of a permutation
Recall how to calculate permutations with repetition (replacement) Learn how to calculate permutations without repetition (without replacement)

3 Permutations What is a permutation? A permutation is an arrangement of objects (numbers, letters, words etc…) in a specific order. Let’s list the possible ways we can pick two numbers from 1, 2 and 3 where repetition is okay. If 𝑛 is the # of possibilities and π‘Ÿ is the # of times we choose, there are… 1, 1 2, 1 3, 1 1, 2 2, 2 3, 2 1, 3 2, 3 3, 3 𝑛 π‘Ÿ permutations.

4 Permutations with repetition
We have seen this before. If we use this formula, we can easily calculate the number of permutations possible. Example Three meetings will take place next week and are possible to happen on any day of the week. It is possible for one, two or all three meetings to take place on a single day. How many possible ways are there for the meetings to take place? 𝑛=7, π‘Ÿ=3, 𝑛 π‘Ÿ β‡’ 7 3 =343 ways

5 36 20 6 8 36 4 81 Months have differing numbers of days. 365.

6 Permutations without repetition
So, permutations with repetition are simple! But without repetition (where a value can’t appear twice) is more complicated. Let’s see the difference. Example (repetition) Example (without repetition) A four digit number is made where zero is allowed in all four positions. How many permutations are there? ? ? ? ? 0 ~ 9; four times 𝑛 π‘Ÿ = 10 4 10 possible 9 possible 8 possible 7 possible =10,000 =10βˆ™9βˆ™8βˆ™7 =5040

7 10βˆ™9βˆ™8βˆ™7…? Factorials! =10βˆ™9βˆ™8βˆ™7βˆ™6βˆ™5βˆ™4βˆ™3βˆ™2βˆ™1 =10! 10! ___________
Does this pattern remind you of anything we learned briefly at the start of this year? =10βˆ™9βˆ™8βˆ™7βˆ™6βˆ™5βˆ™4βˆ™3βˆ™2βˆ™1 =10! How can we represent 10βˆ™9βˆ™8βˆ™7 in terms of factorials? 10! ___________ = 10! 6! 6βˆ™5βˆ™4βˆ™3βˆ™2βˆ™1 How about the general case…with 𝑛 and π‘Ÿ?

8 Permutations without repetition
In our example, we have 10 choices for each digit and there are 4 so… 𝑛= 10 π‘Ÿ= 4 10! 6! The top number is obviously 𝑛! 𝑛! But where does the bottom number come from? (____ = 6!…?) _____ π‘›βˆ’π‘Ÿ ! We clearly did 10 – 4, right? (𝑛 – π‘Ÿ) But we got 6!, not 6…so… 𝑛! π‘›βˆ’π‘Ÿ ! is for permutations without repetition where 𝑛 is the # of possibilities and π‘Ÿ is the # of times we choose.

9 Permutations without repetition
Example How many ways can π‘Ž, 𝑏, 𝑐, 𝑑 be ordered where only 2 letters are used but each can’t be used twice? 𝑛= 4 π‘Ÿ= 2 𝑛! π‘›βˆ’π‘Ÿ ! = 4! 4βˆ’2 ! = 4! 2! = 4βˆ™3βˆ™2βˆ™1 2βˆ™1 = 24 2 =12 12 ways (- π‘Ž, 𝑏 π‘Ž, 𝑐 π‘Ž, 𝑑 𝑏, π‘Ž 𝑏, 𝑐 𝑏, 𝑑 𝑐, π‘Ž 𝑐, 𝑏 𝑐, 𝑑 𝑑, π‘Ž 𝑑, 𝑏 𝑑, 𝑐)

10 Answers 120 72 240 81 210

11 Answersγ€€(Part 3) 64 24 216 120 49 42 256 24 With repetition: 343, Without repetition: 210

12 Answersγ€€(Part 4) 6 30 𝑛=3, π‘Ÿ=3 π‘€π‘–π‘‘β„Žπ‘œπ‘’π‘‘ 12 𝑛! 36 4 8

13 Answers (Part 5) 9P8 2 20 870 720 Crazy: n = 5, r = 5 ⟹5!
Because we can’t pick more options than there are as one is removed each pick. Crazy: n = 5, r = 5 ⟹5! Hello has β€˜l’ twice (indistinguishable) so it will have less than 5!. Or…factorials need to be positive, (n – r)! β‰₯0.

14 Answersγ€€(Part 6) 64 4!=24 720 156 6 7 49! 45! You can’t get 7 different outcomes from a die.


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