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Counting If you go to Baskin-Robbins (31 flavors) and make a triple-scoop ice-cream cone, How many different arrangements can you create (if you allow.

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Presentation on theme: "Counting If you go to Baskin-Robbins (31 flavors) and make a triple-scoop ice-cream cone, How many different arrangements can you create (if you allow."— Presentation transcript:

1 Counting If you go to Baskin-Robbins (31 flavors) and make a triple-scoop ice-cream cone, How many different arrangements can you create (if you allow repeats)? If you don’t allow repeats? = 313 = 29791 = 26970

2 Counting You need to visit 5 cities. How many possible different orderings of visiting the cities (no repeats) are there? This calculation is common enough that we have a notation for it – the factorial (symbolized by !)

3 Permutations and Combinations
These deal with drawing items without replacement Are different orderings counted separately? In other words, is ABC considered different than BAC? If YES, we’re talking about Permutations (arrangements) If NO, we’re talking about Combinations

4 Permutations How many different 4-number PIN numbers are possible if no two numbers are the same? Notice Where’d the 6 come from? From 10-4 = 6

5 Permutations The number of permutations (arrangements or sequences) of r items selected from n available items (without replacement) is:

6 Permutations Suppose you have to pick 5 photos out of 15 for a magazine layout. Ordering matters (different layouts should be considered as separate). How many possible layouts are there?

7 Combinations The number of combinations of r items selected from n different items (without replacement) is: The extra term in the denominator makes it so different arrangements aren’t counted separately (ABC is considered equivalent to CBA)

8 Combinations Suppose you have to pick 4 of your 12 friends to take on a free vacation you won. How many possibilities are there? For comparison’s sake, the number of permutations would be 11880

9 Combinations for Probability
Suppose you have to pick 4 of your 12 friends to take on a free vacation you won. You decide to choose by pulling names from a hat. What is the probability that Ann, Betty, Carlos, and Dean are chosen? That’s one choice out of all possibilities, so

10 Homework 3.7: 1, 3, 5, 9, 21, 25


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