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**Permutations and Combinations**

Making your own starting lineup!

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**Why do we want to study permutation and combination?**

Very often we will find situations where one thing can be done in different ways. In order to find the best way, we need to know how many possible ways are there in total. For example, computer A send message to computer B through the network. How many possible routes are there? Which one is the best? A B

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**Activity 1: Scissor Paper Rock**

Two students play scissor paper rock How many possible ways the play will go? Student 1 Scissor Paper Rock Student 2 Scissor Paper Rock

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Activity 2: Group play Two groups of students. Each time one student from each group play scissor paper rock against each other until all students have played. If you have 4 students in the group, how many different lineups can you possibly make? Student 1 Student 2 Student 3 Student 4 Student 1 Student 3 Student 2 Student 4 Student 4 Student 3 Student 2 Student 1 …, …, …, …, Permutation. 4!=24

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What if you only have 3 students in the group, but you are playing against a group with 4 students? How many different lineups can you possibly make? *each of the 3 students have to play in each of the first 3 rounds, the 4th round you can assign any of the 3 to play again. Student 1 Student 2 Student 3 Student 1 Student 2 Student 3 Student 3 Student 2 Student 1 …, …, …, …, First 3 spots: Permutation. 3!=6 Last spot: 3. Total 6x3=18.

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**Activity 3: Combinations**

A group of n students (n>=3), pick 2 students from them to form a team to play against another team from another group. How many possible ways you can form your team? (Assuming the playing order within a does not matter.) Combination: 4*3/2*1=6 How many possible ways you can form your team? (Assuming the playing order matters.) This goes back to permutation again: 4*3=12

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**4 piles of cards. 2 cards in each pile:**

Activity 4: Combinations and Permutations 4 piles of cards. 2 cards in each pile: Pick one from each pile, how many combinations of cards can you possibly get? Lets count! 4 piles of cards. 4 cards in each pile: Pick 2 cards from each pile, how many combinations of cards can you possibly get? Lets calculate! Each pile has combinations: C(4,2)=6 6x6x6x6=1296

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**4 piles of cards. 2 cards in each pile:**

Pick one from each pile in sequence, how many permutations of cards can you possibly get? Lets count! ♠A, ♥A, ♦A, ♣A. ♠A, ♥A, ♦A, ♣2 ♠A, ♥A, ♦2, ♣A. … 16*4!=384

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**4 piles of cards. 4 cards in each pile:**

Pick 2 cards from each pile in sequence, how many permutations of cards can you possibly get? Lets calculate! Assume the choice of pile sequence is determined. Permutation within each pile: P(4, 2). Permutation of the whole sequence: [P(4, 2)]^4 Now consider the permutation of 4 piles = 4! So total permutation = 4! x [P(4, 2)]^4.

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Statistics 1: Elementary Statistics Section 4-7. Probability Chapter 3 –Section 2: Fundamentals –Section 3: Addition Rule –Section 4: Multiplication Rule.

Statistics 1: Elementary Statistics Section 4-7. Probability Chapter 3 –Section 2: Fundamentals –Section 3: Addition Rule –Section 4: Multiplication Rule.

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