Lecture-3

# Lecture-3 - Lecture 3 Combinatorial Constructions Many...

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Unformatted text preview: Lecture 3. Combinatorial Constructions Many probability spaces arise from combinatorial structures. Permutations. If a numbered set of n items is rearranged in a new order (e.g., 123456 might be rearranged as 352461), there are n possibilities for the first element, n- 1 for the second, n- 2 for the third, etc. Therefore the total number of possible rearrangements is n ! = n · ( n- 1) · ( n- 2) · · · · · 3 · 2 · 1 . Fact 1. There are n ! ways of arranging n distinct elements in order. Consider an experiment consists in rearranging an n-element set randomly in such a way that all rearrangements are equally probable. We can model this with a probability space in which the n ! rearrangements are the outcomes and each has an assigned probability mass of 1 /n !. For example, after many shuffles of a deck of cards there is no rearrangement that is more probable than any other. Thus, each card sequence has a probability of 1 / 52!. (52! is about 8 × 10 67 .) If k elements are taken from an n-element set in order, without replacement, there are n possibilities for the first choice, n- 1 for the second, n- 2 for the third, . . . and n- k +1 for the k th choice. Therefore, the total number of ways of completing this task is n ! = n · ( n- 1) · ( n- 2) · · · · · ( n- k + 1) = n ! ( n- k )! . Fact 2. There are n ! ( n- k )! ways of selecting k elements in order from a collection of n distinct elements....
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## This note was uploaded on 11/29/2011 for the course MATH 3355 taught by Professor Britt during the Spring '08 term at LSU.

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Lecture-3 - Lecture 3 Combinatorial Constructions Many...

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