Permutations and Combinations

Plan

• homework update and reminder

This time:

• charts and graphs
• birthday paradox
• poker probabilities

If there's time, we'll talk about the connection between "choose," Pascal's triangle, and the binomial coefficients. This connection can be derived in many ways, but one way is to rely on a recursive definition of (n choose k):

(n choose k) = (n-1 choose k-1) + (n-1 choose k)

The coefficient of a^kb^n-k is (n choose k) because when we compute

(a+b)ⁿ = (a+b)(a+b)^n-1

all terms of the form a^kb^n-k are found in one of two ways

• "a" multiplied by the term a^k-1b^n-k, and there are (n-1 choose k-1) of these
• "b" multiplied by the term a^kb^n-k-1, and there are (n-1 choose k) of these

Here's a link to Dr. Math's explanation of Pascals' triangle, which may be useful.

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