Counting Rules

1. Counting Rules

2. The probability of a specific event or outcome is a fraction. In the numerator we have the number of ways the specific event can occur and in the denominator we have the total number of possible outcomes. Often we can see the numbers fairly easily in a problem. Sometimes we can’t. Counting rules are useful when we can not easily see the number of ways things can occur. We will consider 5 counting rules.

3. Rule 1 If any one of k different mutually exclusive and collectively exhaustive events can occur on each of n trials, the number of possible outcomes is equal to kn (k raised to the nth power). Sometimes this will be written as k^n, where ^ means the next number should be treated as a power. Example: flip a coin (k = 2) and heads or tails are mutually exclusive (only one can occur on a flip) and collectively exhaustive (either heads or tails must occur). If you flip the coin three times one outcome might be HHT. Another outcome might be THT. The total possible number of outcomes would be 2^3 = 2 times 2 times 2 = 8.

4. Coin example All eight outcomes when you flip a coin 3 times (or just flip three coins once) are: HHH, HHT, HTH, HTT, TTT, TTH, THT, THH. So, what is the probability you would get 2 heads when you flip a coin 3 times? 2 heads occurs 3 out of the 8 times for a probability of 3/8 = .375.

5. Rule 2 Multiple-Step Experiments If an experiment can be described as a sequence of n steps or trials with k1 possible outcomes or events on the first step, k2 possible outcomes on the second step, and so on until we get to kn, then the total number of experimental outcomes is the product (k1)(k2)...(kn). Say a construction project has two stages - design and construction. If the design could be completed in 2, 3, or 4 months and the construction could be done in 6, 7, or 8 months, then there are (3)(3) = 9 different experimental outcomes. I will list the outcomes as ordered pairs of numbers, with the first number the time to complete the design and the second the time to complete the construction: (2, 6), (2, 7), (2, 8), (3, 6), (3, 7), (3, 8), (4, 6), (4, 7), and (4, 8).

6. Example continued What is the probability the project will be completed in 9 months? 2 of the 9 outcomes would result in completion in 9 months for a probability of 2/9 = .22

7. Rule 3 The number of ways that n items can be arranged in order is n! = n times (n – 1) times (n – 2) times … (1). n! is called n factorial and 0! = 1 by definition. Example: say you have 3 letters, A, B, and C. How many ways can you arrange the letters? 3! = 3 times 2 times 1 = 6. The six ways are ABC, ACB, BCA, BAC, CAB, and CBA. What is the probability your arrangement will end in A? 2/6 = .33

8. Examples In baseball at the start of the game the manager presents the starting line-up. Part of this is the batting order (the other part is the position in the field of each player). If you have nine players in the starting line-up, how many different batting orders are possible? You could have (9)(8)(7)(6)(5)(4)(3)(2)(1) = 362,880 batting orders. If you have 6 books, how many ways could you order them on a shelf? You could have (6)(5)(4)(3)(2)(1) = 720 arrangements.

9. Rule 4 Permutations - Remember having a lock at school? The dial on the lock might have had 40 numbers. To open the lock you spun the dial to the right several times and settled on the first number in the combination, then you went around to the left once past that number and then settled on the second number, then you went right to the third number. Say you had the combination 7 - 16 - 32. This is 3 numbers from 40. Is the combination 32 - 7 - 16 the same as 7 - 16 -32? The answer is no. Parker, what is the point? The idea of a combination lock means 7 - 16 - 32 and 32 - 7 - 16 would be different combinations. This is really the idea of permutation! Perhaps a better name for our locks would be permutation locks – order matters.

10. The formula for the number of permutations when arranging X objects selected from n objects is n!/(n- X)!. For example 2 from 5 is 5! = (5)(4)(3)(2)(1) = (5)(4) = 20 (5 - 2)! (3)(2)(1) Let’s do another example. Say we have the letters A, B, and C. Say we want to choose 2 of these. We have the number of permutations 3!/(3-2)! = 3(2)/1 = 6. The permutations would be AB, BA, AC, CA, BC, CB.

11. If you have 6 books you can arrange them 720 ways on the shelf. In general, with n items you can order them n! ways. (that reminds me of a bad joke – what’s a henway? Oh, about 5 pounds) If you only have room for 4 of the 6 books you can arrange them 6!/(6 – 4)! = 6!/2! = (6) (5)(4)(3)(2)(1)/(2)(1) = (6)(5)(4)(3) = 360 ways. What is going on here? 6 books can be arranged 720 ways but if you only have room for 4 you have 360 arrangements possible. Note in the denominator of the permutation we have the 2 books not on the shelf in factorial form. What this really means (say I have books 1 through 6) is that you could have books 1, 2, 3, 4 on the shelf, 2, 3, 4, 5 on the shelf and so on. All six would be rotated in, but 4 at a time.

12. Rule 5 Combinations – The formula for the number of combinations when taking X objects from n objects, irrespective of order is n!/[X!(n - X)!]. For example 2 from 5 has combinations 5! = (5)(4)(3)(2)(1) = (5)(4) = 10 2!(5 - 2)! 2!(3)(2)(1) 2 Let’s do another example. Say we have the letters A, B, and C. Say we want to choose 2 of these. We have the number of combinations 3!/2!(3-2)! = 3(2)(1)/2(1)(1) = 3. The permutations were AB, BA, AC, CA, BC, CB, but in a combination since order does not matter AB and BA are the same and only counted once (and similarly for AC and BC).

13. Say you have an ordinary deck of playing cards - you know, ace through king in spades, hearts, diamonds, and clubs. So there are 52 cards in the deck. In many games of poker you get 5 cards. How many different combinations of 5 cards are there? 52! = (52)(51)(50)(49)(48) = 2598960 5!(47!) (5)(4)(3)(2)(1) This means there are 2 million, 598 thousand, 960 different combinations of hands you could be dealt. Most hands you get are not memorable - you know, you get a 7 of hearts, queen of spades, 3 of clubs, 9 of clubs and a 4 of spades. But a royal flush hearts - 10, jack, queen, king, and ace all hearts - is memorable. Each hand mentioned has a 1 divided by 2598960 chance of happening. But the royal flush is a winner!

14. Note on the previous screen I had 52! = (52)(51)(50)(49)(48) = 2598960. 5!(47!) (5)(4)(3)(2)(1) Often with a combination or permutation calculation we can cancel out a lot of terms in the numerator and the denominator. I really should have put 52! = (52)(51)(50)(49)(48)(47)(46)(45)…(1) = 2598960 5!(47!) (5)(4)(3)(2)(1)(47)(46)(45)…(1), But then we cancel 52! = (52)(51)(50)(49)(48)(47)(46)(45)…(1) = 2598960 5!(47!) (5)(4)(3)(2)(1)(47)(46)(45)…(1), To get 52! = (52)(51)(50)(49)(48) = 2598960. 5!(47!) (5)(4)(3)(2)(1)

15. Let’s go back to the example with 6 books. There are n! = 6! = 720 ways of arranging all 6 books in order, n!/(n – X) = 6!/2!= 360 or arranging them in order 4 at a time, and n!/[X!(n - X)!] = 6!/(4!2!) = 15 ways of having 4 of the 6 on the shelf at any one time without the order mattering.