 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
Permutations with Repetition
There are 2 kinds of permutations:
Permutations with Repetition  You can reuse the same element within the order, such as in the lock from the previous question, where the code could be "000".
Permutations without repetition  Each element can only appear once in the order. For example, on some locks to houses, each number can only be used once. Similarly, when you're ranking people in the poetry contest, each slot needs to be given to a different person.
Calculating Permutations with Repetition
Here's a question to ponder: Let's say you had a simple lock like with just 2 dials and 3 digits per dial. A code consists of 2 digits in the correct order. How many possible codes could it have?
(Scroll down for the solution.)
Permutation Tree
This tree plots out the different possibile codes for the lock:
For each of the 3 possibilities of the first dial, there are 3 possibilities for the second dial. This means there will be 3x3 = 9 possibilities in total. Whenever there are n ways of doing one thing and m ways of doing another, there are n * m ways of doing both together.
So if you added a 3^{rd} dial, then there would be 9*3 = 27 possible permutations of all 3 dials. This is because for each of the 9 permutations of the 1^{st} two dials, there will be 3 possibilities for the 3rd dial.
Formula for Permutations with Repetition
If there are N possibilities for each element, and there are E elements, then:
Total # of Possibilities = N * N * N * ... (E times)
This can also be written as an exponent, as N^{E}.
Challenge
How many codes are possible for the 3dial lock with 10 digits (09) on each dial? (Enter a number answer.)
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Comments
Chillex
Aug 9, 10:25 PMI got a little confused about N as possibilities and E as Dials. But I still got the correct answer.
rodrigo perez
Sep 28, 5:50 PMThere are 10 ways to set each of the 3 dials... so the first combination would be 000, and the last one, 999.
Ian Rainford
Nov 15, 6:41 PMwhy is 10^{3} incorrect?
Learneroo
Nov 16, 6:35 PM@Ian, in this question you need to enter the actual answer.