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Permutations (11-8)

Permutations

Vocabulary

Permutations-

Factorial-

Examples

A formula for the number of possible permutations of k objects from a set of n. This is usually written nPk .

Formula:

Example:

How many ways can 4 students from a group of 15 be lined up for a photograph?
Answer: There are 15P4 possible permutations of 4 students from a group of 15.


different lineups


Here’s an easy way to remember: permutation sounds complicated, doesn’t it? And it is. With permutations, every little detail matters. Alice, Bob and Charlie is different from Charlie, Bob and Alice (insert your friends’ names here).
Combinations, on the other hand, are pretty easy going. The details don’t matter. Alice, Bob and Charlie is the same as Charlie, Bob and Alice.
Permutations are for lists order matters) and combinations are for groups (order doesn’t matter).
A joke: A "combination lock" should really be called a "permutation lock". The order you put the numbers in matters. (A true "combination lock" would accept both 10-17-23 and 23-17-10 as correct.)

Permutations: The hairy details

Let’s start with permutations, or all possible ways of doing something. We’re using the fancy-pants term “permutation”, so we’re going to care about every last detail, including the order of each item. Let’s say we have 8 people:
1: Alice
2: Bob
3: Charlie
4: David
5: Eve
6: Frank
7: George
8: Horatio
How many ways can we award a 1st, 2nd and 3rd place prize among eight contestants? (Gold / Silver / Bronze)
permuation_medals.png
We’re going to use permutations since the order we hand out these medals matters. Here’s how it breaks down:
  • Gold medal: 8 choices: A B C D E F G H (Clever how I made the names match up with letters, eh?). Let’s say A wins the Gold.
  • Silver medal: 7 choices: B C D E F G H. Let’s say B wins the silver.
  • Bronze medal: 6 choices: C D E F G H. Let’s say… C wins the bronze.
We picked certain people to win, but the details don’t matter: we had 8 choices at first, then 7, then 6. The total number of options was 8 * 7 * 6 = 336.
Let’s look at the details. We had to order 3 people out of 8. To do this, we started with all options (8) then took them away one at a time (7, then 6) until we ran out of medals.
We know the factorial is: \displaystyle{8! = 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 }
Unfortunately, that does too much! We only want 8 * 7 * 6. How can we “stop” the factorial at 5?
This is where permutations get cool: notice how we want to get rid of 54321. What’s another name for this? 5 factorial!
So, if we do 8!/5! we get:
\displaystyle{\frac{8!}{5!} = \frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}  = 8 \cdot 7 \cdot 6}
And why did we use the number 5? Because it was left over after we picked 3 medals from 8. So, a better way to write this would be:
\displaystyle{\frac{8!}{(8-3)!}}
where 8!/(8-3)! is just a fancy way of saying “Use the first 3 numbers of 8!”. If we have n items total and want to pick k in a certain order, we get:
\displaystyle{\frac{n!}{(n-k)!}}
And this is the fancy permutation formula: You have n items and want to find the number of ways k items can be ordered:


\displaystyle{P(n,k) = \frac{n!}{(n-k)!}}

http://betterexplained.com/articles/easy-permutations-and-combinations/

Video Help

https://www.khanacademy.org/math/precalculus/prob_comb/combinatorics_precalc/v/permutation-formula

https://www.youtube.com/watch?v=-EEDAsAF1lg

https://www.youtube.com/watch?v=CxBT2m9roh4

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