Lecture Week 7

# Discrete Mathematics and Its Applications with MathZone

• Notes
• davidvictor
• 19

This preview shows pages 1–5. Sign up to view the full content.

Permutations r-permutation (AKA “ordered r-selection”) An ordered arrangement of r elements of a set of n distinct elements. permutation of a set of n element = n-permutation on this set Example: S={J,K,Q}: (K,Q,J) is a permutation of S; (J,K) is a 2-permutation of S Note: Set is unordered , but permutation is ordered !, e.g. (K,Q)≠(Q,K) The number of r-permutations of n objects is: P(n,r) =n(n-1)(n-2)...(n-r+1) =n! / (n-r)! 0<=r<=n And the number of permutations of n objects is P(n,n) = n! (Proof:1 st object can be chosen in n ways, 2 nd in (n-1) ways, and so on (k th in n-k+1 ways) until the r th object is chosen in n-r+1 ways. Now, use the product rule to get the formula given for P(n,r).) Reminder: n! = n(n-1)(n-2)...1. “n factorial.” (n+1)! = (n+1)n! 0! = 1

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

Permutation Examples Q: A mailman needs to bring 8 packages to 8 cities. How many ways are there to visit the cities? A: Pick first city among 8, second city among 7, etc. Each route is a permutation of 8-element set: Answer = P(8)=8! Q: How many permutations of the letters “abcdefgh” contain “abc” as a block. A: Rename “abc” to B. Question becomes: Count permutations of blocks Bdefgh: A: It’s the # permutations in a 6-element set, i.e. P(6) = 6! Q: How many ways to pick three prizes (gold,silver,bronze) among 100 contestants? A: It’s the number of 3-permutations in a 100-element set: P(100,3) = 100! / (100-3)! = 100! / 97! = 100*99*98
Combinations r-combination (AKA “ un ordered r-selection”) An unordered selection of r elements, i.e. a subset of size r of a set of n elements. Example: S={A,J,Q,K}. {A,J,K}={K,A,J}={J,K,A} are all 3-combinations of set S. The total number of r-combinations of a set of size n is denoted C(n,r), and given by: C(n,r) = n! / (r! (n-r)!) , for 0<=r<=n Two facts to observe: 1. C(n,r) = C(n,n-r) Why does this make sense? 2. C(n,r) = P(n,r) / r! Why? Each of the C(n,r) subsets of r objects can be ordered in r! ways, meaning that C(n,r) * r! = P(n,r)

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

Computing Combinations Computing C(n,r) as n! / [(n-r)!r!] is grossly inefficient and, as a practical matter, can lead to incorrect results (even using computers!).
This is the end of the preview. Sign up to access the rest of the document.
• Fall '07
• Jarecki
• ways

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern