 Intro to Combinations and Permutations
 Permutations with Repetition
 Permutations without Repetition
 Permutations Formula

License Plates  Android Lock Screen
 Tigers and Bears  Oh My

Combination without Repetition  Combination Formula
 Combining Combinations

The String of Beads  Anagrams
 Combinations with Repetition I
 Combinations with Repetition II
Anagrams
Let's say you were given 1 Plum, 2 Oranges, and 1 Lime to put on a string. How many ways could you arrange them?
You could represent the fruit as P, O, O and L and create a table to show all the possibilities:
P  L  O  O 

POOL  LOOP  OLOP  OPOL 
POLO  LOPO  OLPO  OPLO 
PLOO  LPOO  OOLP  OOPL 
There are 12 possible arrangements, or permutations.
Q: You can't repeat any number of fruit/characters in your arrangement, but according to the node permutations without repetition, 4 characters should be able to be arranged in 4! = 24 possible ways, not 12.
Question: How could you solve this problem in general?
Try to figure out the solution before looking ahead, perhaps by seeing how permutations could be created by "POOOL".
Solution
Here are 2 possible approaches you can take to this problem:
Convert to recent problem
In the arranging beads problem, permuting black and white beads was really just a problem of choosing which beads should be black. You can do the same thing with 3 and more types. In this example, you can choose which spots will be 'O' 'P' and 'L':
O  5 Choose 3 = 10. You will then have 2 remaining spots to choose from, so:
P  2 choose 1 = 2. Finally, there's 1 choice left:
L  1 choose 1 = 1.
Since you will need to choose where to put the P's for each possible 'O' choice, you multiply the different choices together to get the total:
10 * 2 * 1 = 20 possible permutations of POOOL.
Direct Approach
In the word POOOL, if you use the regular formula (5!), you will get duplicates from the triple O's. How many duplicates? The 3 O's could have been arranged in 3! different ways (if they had been different), so for each word with 3 O's there will be 3! 'false' ways to arrange the O's. The same technique is used to convert a permutation to a combination.
So 5! / 3 != 20 possible permutations of POOOL.
General Formula
If a group G contains n elements, and r of the elements are identical, then the number of unique permutations of G equals:
If there are additional duplicated characters, which are duplicated r_{1}, r_{2}... times, then the formula will be:
n!
r_{1}!, r_{2}!, etc...
Challenge
How many ways can you arrange all the letters in the word "BANANA"?
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