Combinatorics

Combinatorics is closely related to probability in cases where each combination is equally likely.

  • if each combination/permutation is equally likely, then the odds of landing upon any particular set is equal to the total possible number of sets.

When determining the result of a combinatorics question, you must ask:

  1. does order of the result set matter?
  2. can the same value be used multiple times?
    • if yes, it's a (permutation/combination) with repetition
    • if no, it's a (permutation/combination) without repetition

Each combination of the 4 possibilities will yield a different formula.

The Basic Counting Principle

When there are mm ways to do one thing, and nn ways to do another, then there are m×nm×n ways of doing both.

  • ex. there are 3 girls and 4 boys, meaning there can be 12 possible pairs.
  • ex. you have 3 shirts and 4 pants, meaning 3×4=123×4=12 different outfits.
  • ex. a password consists of 2 letters followed by 3 numbers (a permutation with repetition). To get the number of total possible passwords, we have to get the # of possibilities of the letter part and multiply it by the # of possibilities for the number part. (answer: 676,000)
  • ex. I want to buy a car. It can be any combination of:
    • type: sedan, hatchback
    • color: red, blue, green
    • trim: standard, sports, luxury
    • Therefore, 2×3×3=182 \times 3 \times 3 = 18 possible combinations
  • this formula only works when choices are independent of each other (such as when we have the logic "black hatchbacks are not possible")

Handshake problem

If there are 7 people at a party and each of them shake hands, how many handshakes were there?

handshakes=n(n1)2handshakes = n*{(n - 1)\over2}

This is because each of the n people can shake hands with n - 1 people (they would not shake their own hand, which is why we divide by 2), and the handshake between two people is not counted twice.

Chess board combinations

If there are 6 players and 3 chess boards, how many different combinations of players can there be?

  • 2 players at the same chess board switching places (ie. black/white) counts as 1 combination
  • it does not matter which board the same pair of players sits at.
combinations=(n1)×rcombinations = (n - 1) \times r
combinations=5×3=15combinations = 5 \times 3 = 15

Or:

n!2r×nn! \over 2^r \times n
6!23×6=15{6! \over 2^3 \times 6} = 15

Children
  1. Combinations
  2. Permutation

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