Calculating the number of permutations or combinations.
Understand the difference between a combination and a permutation
Be able to calculate the number of permutations with and without repetition
Be able to calculate the number of combinations without repetition
Calculating the number of different possible outcomes is called combinatorics.
To better understand this concept, lets start off by looking at an example.
To better understand this concept, we need to look at all the parts that make it up and see how they all interact with one another.
1. Counting Principle
The best way to understand the counting principle and how it works,is by doing it through practice.
You're planning a date: dinner, entertainment and dessert.
You have two choices for dinner, Happy Meals at McDonald's and microwave burritos at the local Quickymart.
You have three choices for entertainment: bowling, a movie or watching wrestling on TV.
You have two choices for dessert: s'mores and pie. How many possible dates are there?
When in doubt make a tree diagram
2. Factorials (!)
If we have 5 boys and 5 chairs in a row, and want to find out how many different ways the boys can arrange themselves in the chairs, we have 5 choices of who can sit in the first chair, 4 choices of who can sit in the second chair, then 3, 2 and 1, so we have 5 x 4 x 3 x 2 x 1 possible orders.
4. Combinations
Combinations are trickier.
Figure out the number of permutations, then figure out how many times each correct answer (each combination) is counted, and divide by that number.
If we are trying to choose 4 students out of a class of 10 for a MATHCOUNTS team, how many teams can we make?
Well, we have 10 choices of who to chose first, 9 choices of who to chose second, then 8 and 7.
So we have 10 x 9 x 8 x 7 possibilities.
But wait, it doesn’t matter who we chose first, second, third or fourth. So Andrea, Beth, Carrie, David is just the same as Beth, David, Carrie, Andrea.
How many different ways would we have chosen this same team?
We have 4 choices of who we picked first
3 choices of who we picked second.
2 choices of who we picked third and
1 choice of who we pick fourth.
So each team is included 4 x 3 x 2 x 1 times.
So our answer is (10 x 9 x 8 x 7) / (4 x 3 x 2 x 1) or 210. (Fortunately, that works out the same as doing it the “standard” way of using the formula for 10C4 “10 choose 4”.)
Here's the official formula:
Mathmom needs to choose 4 students to be on the MATHCOUNTS team. There are twenty students she can choose among (in her dreams!). How many different teams can she make?
A certain lottery is played by choosing your own set of 6 winning numbers from among the numbers 1 through 49. How many possible such combinations are there?
There are 12 boys and 14 girls in Ms. Brown’s class. She needs to choose 3 boys and 3 girls for a debate team. How many different teams are possible?
(a) In a standard deck of 52 cards, how many different 7 card hands are possible?(b) If you separate the cards into suits and keep only the hearts, how many different collections of 5 hearts are possible?
Betty is about to order dinner at her favorite restaurant. She will order a drink, an appetizer, a main course, 2 different side items, and a dessert. If there are 10 choices for drinks, 5 appetizers, 6 main courses, 8 side items, and 5 desserts, in how many ways can Betty order her meal?
http://webbapp.liber.se/matematikboken-z/#
http://webbapp.liber.se/matematikboken-z/#/5-sannolikhet-statistik
Textbook Practice
Ett pg. 240 no. 5053 ab, 5054.
Två pg. 240 no. 5058, 5059.
Tre pg. 241 no. 5061 ab, 5062, 5063.
Fyra pg. 241 no. 5065, 5066, 5067.