VIEWS: 141 PAGES: 15 POSTED ON: 6/17/2011
COUNTING TECHNIQUES PERMUTATIONS AND COMBINATIONS Computer Science, Statistics and Probability all involve counting techniques which are a branch of mathematics called combinatorics (ways to combine things). For dinner you have the following choices: APPETIZERS MAINS soup salad chicken shrimp hamburger DESSERTS How many different combinations of meals could you make? We'll build a tree diagram to ice cream show all of the choices. Notice the number of choices at each branch We ended up with 12 possibilities 2 3 2 choices choices choices soup, chicken, ice cream 2 3 2 = 12 soup, chicken, shrimp soup, shrimp, ice cream soup, shrimp, soup, hamburger, ice cream soup, hamburger, salad, chicken, ice cream salad, chicken, shrimp salad, shrimp, ice cream salad, shrimp, Now to get all salad, hamburger, ice cream possible choices we salad, hamburger, follow each path. Multiplication Principle of Counting If a task consists of a sequence of choices in which there are p selections for the first choice, q selections for the second choice, r selections for the third choice, and so on, then the task of making these selections can be done in pqr different ways. If we have 6 different shirts, 4 different pants, 5 different pairs of socks and 3 different pairs of shoes, how many different outfits could we wear? 6 4 5 3 = 360 A permutation is an ordered arrangement of r objects chosen from n objects. For combinations order does not matter but for permutations it does. There are three types of permutations. The first is distinct with repetition. This means there are n distinct objects but in this means different choosing r of them you can repeat an object. Let's look at a 3 There are 10 choices for the first number combination lock There are 10 choices for the second number with numbers 0 and you can repeat the first number through 9 There are 10 choices for the third number and you can repeat By the multiplication principle there are 10 10 10 = 1000 choices This can be generalized as: Permutations: Distinct Objects with Repetition The number of ordered arrangements of r objects chosen from n objects, in which the n objects are distinct and repetition is allowed, is nr What if the lock had four choices for numbers instead of three? 104 = 10 000 choices The second type of permutation is distinct, without repetition. Let's say four people have a race. Let's look at the possibilities of how they could place. Once a person has been listed in a place, you can't use that person again (no repetition). First place would be Based on the multiplication principle: choosing someone from 4 3 2 1 = 24 choices among 4 people. Now there are only 3 to choose from for second place. Now there are only 2 to choose from for third place. 4th 3rd 2nd 1st Only one possibility for fourth place. nPr, means the number of ordered arrangements of r objects chosen from n distinct objects and repetition is not allowed. n! nP r n r ! 4! 4 3 2 1 4P 24 In the last example: 4 4 4 ! 0! If you have 10 people racing and only 1st, 2nd and 3rd place how many 0! = 1 possible outcomes are there? 10! 10 9 8 7! 10 P 720 3 10 3 ! 7! A combination is an arrangement of r objects chosen from n objects regardless of order. nCr , means the number combinations of r objects chosen from n distinct objects and repetition is not allowed. n! n n Cr or n r ! r! r Order doesn't matter here so the combination 1, 2, 3 is not different than 3, 2, 1 because they both contain the same numbers. You need 2 people on your committee and you have 5 to choose from. You can see that this is without repetition because you can only choose a person once, and order doesn’t matter. You need 2 committee members but it doesn't matter who is chosen first. How many combinations are there? 5! 5 4 3! 5 C2 10 5 2 !2! 3!2! The third type of permutation is involving n objects that are not distinct. How many different combinations of letters in specific order (but not necessarily English words) can be formed using ALL the letters in the word REARRANGE? E R N R A A G E R The "words" we form will have 9 letters so we need 9 spots to place the letters. Notice some of the letters repeat. We need to use R 3 times, A 2 times, E 2 times and N and G once. 9C3 6C2 4C2 2C1 1C1 84 15 6 2 1 = 15 120 possible "words" First we choose That leaves That leaves positions for the 6 positions 2 positions R's. There are 9 That leaves for 2 A's. That leaves for the N. positions and we'll 1 position choose 3, order 4 positions for 2 E's. for the G. doesn't matter This can be generalized into the following: Permutations Involving n Objects That Are Not Distinct The number of permutations of n objects of which n1 are of one kind, n2 are of a second kind, . . ., and nk are of a kth kind is given by n! n1 ! n2 ! nk ! where n n1 n2 nk A national hamburger chain used to advertise that it fixed its hamburgers “256 ways” since customers could order whatever toppings they wanted. How many toppings must have been available? For each topping, the choice is yes (put it on my burger!) or no (leave it off!). 2n = 256 log 2n = log 256 n log 2 = log 256 log256 n log2 n=8 There must have been 8 possible toppings. Group Assignment For each of the following numbers, make up a counting problem that has the number as its answer. 1. 52C3 2. 12C3 3. 25P11 4. 25 Homework • Pg 609 – 611 #11, 13, 17, 19, 25, 35, 39, 45, 51