Just another way of thinking of combinatorics.

• Permutations: ALL possible ways of doing something (order matters)
  • What are all the permutations of some elements?
  • How many ways to pick a gold, silver and bronze medalist from 8 runners?
    • First choice: 8 possible
    • Second choice: 7
    • Third: 6
    • Answer = 8 * 7 * 6
    • 8!/(8-3)! is just a fancy way of saying "Use the first 3 numbers of 8!"
  • n!/(n-k)! is just "Use the first k numbers of n!"
• Combination: ONE possible way of doing something (order doesn't matter - different orders are the same answer)
  • How many unique ways can you choose 3 people from 8?
  • The order we pick them doesn't matter. Picking Alice, Bob and Charlie is ONE answer (it is that same as Bob, Alice, and Charlie).
  • Answer: total number of ways / redundancies
    • Total number of ways is just 8 * 7 * 6, a permutation
    • The number of redudancies is interesting. How many ways can we rearrange 3 people?
      • 3 choices for first
      • 2 choices for second
      • 1 choice for thirt
      • 3 * 2 * 1 = 3! = 6
    • Number of redundancies = k!
    • Interestingly, we find the number of redundancies using a permutation :)
• n!/k!(n-k)! is just "Use the first k numbers of n!, and divide by the k! variants of each answer"