Permutations And CombinationsInduction

Permutations And CombinationsInduction - 1 Permutations And...

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

View Full Document Right Arrow Icon
1 Permutations And Combinations: Proof by Induction We have the relation expressed by the arrows can be written mathematically as: 1 1 n n n k k k           We could also derive this formula by considering the definition of “n choose k” as the number of subsets. When we have n+1 items to choose from, we can pick out one of the items, call it A, and then say: every subset having exactly k items either does contain A, or does not contain A. If it does not contain A, then it is one of the subsets that contains k items, and is drawn from the set of exactly n items that we have when we do not contain A. There are exactly n k    of these subsets. OR, it does contain the element A. It must also have 1 k other elements, and they must be drawn from the 1 n other elements in the set. So the number of such subsets (the ones that do contain A) is exactly 1 n k . The subsets that we count in the first way are completely different from the ones we count in the second way. So the
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 02/20/2012 for the course 790 373 taught by Professor Boros during the Fall '09 term at Rutgers.

Page1 / 2

Permutations And CombinationsInduction - 1 Permutations And...

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