Section+4.6

**§4.6- Permutations**
Today in class we looked at an example question which was a slight variation on one we had previously done. The question we looked at was:

**Suppose the Kindergarteners are gluing 4 shapes in a line and they have 6 options of shapes (squares, circles, stars, ovals, triangles, and diamonds). How many different designs are possible?**
__First way to think about it__: 1st shape: 6 options 2nd shape: 5 options 3rd shape: 4 options 4th shape: 3 options Total number of different designs: 6 x 5 x 4 x 3= 360 This is equal to media type="custom" key="11772884" because you are dividing out the pieces (2!) that you don't use from the total number of possible designs.

__Second way to think about it__: If you arrange all 6 objects there are 6! ways to do it. It seems you can just lop of the last 2 which makes it 6! strings of 4, but this answer is too big. The reason for this is you have some repeats and are double counting if you do this. Since you double count some, you must divide out the 2! to account for this. Therefore the answer you get is still media type="custom" key="11773162"= 360

__Difference between the two methods:__ You divide out the 2! in the second way since you double counted some, but it is simply used to simplify notation in the first.


 * Definition of permutation:** A one-to-one function from a set to itself. There are n! permutations of an n element set.
 * nPk**: The number of ways you can pick k things in order from a set of n distinct things

Relating this back to the problem: If you are ordering k elements of an n element set then there are k slots (__1st slot__, __2nd slot__, __3rd, slot__..__kth slot__).

__n__ __n-1__ __n-2__ __n-3__... __n-k+1__

In the problem we did there were 4 elements picked out of a 6 element set (k=4, n=6).

__6__ __5__ __4__ __3__

__Definitions:__ math $ \frac{n!}{(n-k)!} = _{n}P_{k} $ math

nPk: n(n-1)(n-2)(n-3)...(n-k+1)

nPn= n!

0!=1

math $ _{n}P_{n}=\frac{n!}{(n-n)!} = \frac{n!}{0!}=n! $ math __Examples:__
 * An anagram is the rearrangement of the letters in a word or phrase. How many possible anagrams are there for the following words: **

1) TRIANGLE Answer: There are 8 letters being arranged into 8 positions so the answer is 8!= 40,320

2) PEEL Since the E's repeat, it makes this one a little trickier. To think of it better, make the E's into two different letters: P E E L When you do this you see that it is the same idea of double counting the possible arrangements as in the first problem we did since now there are 4 spaces and 3 different letters. The answer is media type="custom" key="11773256"=12

3) GU L L IB L E Now there are 3 L's so if you think of them all as different, you see that it is the same as triple counting so you must divide out 3! from 8! to get the answer is media type="custom" key="11773284"=6,720