Lecture-3

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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

Page1 / 3

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online