Fundamental Counting Principles and Probability
- 1. Copyright © Cengage Learning. All rights reserved. 8.5 Counting Principles
- 2. 2 What You Should Learn • Solve simple counting problems • Use the Fundamental Counting Principle to solve more complicated counting problems • Use permutations to solve counting problems • Use combinations to solve counting problems
- 4. 4 Example 1 – Selecting Pairs of Numbers at Random Eight pieces of paper are numbered from 1 to 8 and placed in a box. One piece of paper is drawn from the box, its number is written down, and the piece of paper is returned to the box. Then, a second piece of paper is drawn from the box, and its number is written down. Finally, the two numbers are added together. In how many different ways can a sum of 12 be obtained?
- 5. 5 Example 1 – Solution To solve this problem, count the number of different ways that a sum of 12 can be obtained using two numbers from 1 to 8. First number Second number From this list, you can see that a sum of 12 can occur in five different ways.
- 6. 6 The Fundamental Counting Principle
- 7. 7 The Fundamental Counting Principle
- 8. 8 Example 3 – Using the Fundamental Counting Principle How many different pairs of letters from the English alphabet are possible? Solution: There are two events in this situation. The first event is the choice of the first letter, and the second event is the choice of the second letter. Because the English alphabet contains 26 letters, it follows that the number of two-letter pairs is 26 26 = 676.
- 10. 10 Permutations One important application of the Fundamental Counting Principle is in determining the number of ways that n elements can be arranged (in order). An ordering of elements is called a permutation of the elements.
- 11. 11 Example 5 – Finding the Number of Permutations of n Elements How many permutations of the following letters are possible? A B C D E F Solution: Consider the following reasoning. First position: Any of the six letters Second position: Any of the remaining five letters Third position: Any of the remaining four letters Fourth position: Any of the remaining three letters Fifth position: Either of the remaining two letters Sixth position: The one remaining letter
- 12. 12 Example 5 – Solution So, the numbers of choices for the six positions are as follows. The total number of permutations of the six letters is 6! = 6 5 4 3 2 1 = 720. cont’d
- 13. 13 Permutations
- 14. 14 Permutations Using this formula, find the number of permutations of eight horses taken three at a time is = 336
- 15. 15 Permutations
- 16. 16 Example 7 – Distinguishable Permutations In how many distinguishable ways can the letters in BANANA be written? Solution: This word has six letters, of which three are A’s, two are N’s, and one is a B. So, the number of distinguishable ways in which the letters can be written is = 60.
- 17. 17 Example 7 – Solution The 60 different distinguishable permutations are as follows. AAABNN AAANBN AAANNB AABANN AABNAN AABNNA AANABN AANANB AANBAN AANBNA AANNAB AANNBA ABAANN ABANAN ABANNA ABNAAN ABNANA ABNNAA ANAABN ANAANB ANABAN ANABNA ANANAB ANANBA ANBAAN ANBANA ANBNAA ANNAAB cont’d
- 18. 18 Example 7 – Solution ANNABA ANNBAA BAAANN BAANAN BAANNA BANAAN BANANA BANNAA BNAAAN BNAANA BNANAA BNNAAA NAAABN NAAANB NAABAN NAABNA NAANAB NAANBA NABAAN NABANA NABNAA NANAAB NANABA NANBAA NBAAAN NBAANA NBANAA NBNAAA NNAAAB NNAABA NNABAA NNBAAA cont’d
- 19. 19 Combinations
- 20. 20 Combinations When you count the number of possible permutations of a set of elements, order is important. As a final topic in this section, you will look at a method for selecting subsets of a larger set in which order is not important. Such subsets are called combinations of n elements taken r at a time. For instance, the combinations {A, B, C} and {B, A, C} are equivalent because both sets contain the same three elements, and the order in which the elements are listed is not important.
- 21. 21 Combinations So, you would count only one of the two sets. A common example of a combination is a card game in which the player is free to reorder the cards after they have been dealt.
- 22. 22 Example 8 – Combinations of n Elements Taken r at a Time a. In how many different ways can three letters be chosen from the letters A, B, C, D, and E? (The order of the three letters is not important.) b. A standard poker hand consists of five cards dealt from a deck of 52. How many different poker hands are possible? (After the cards are dealt, the player may reorder them, so order is not important.)
- 23. 23 Example 8 – Solution a. You can find the number of different ways in which the letters can be chosen by using the formula for the number of combinations of five elements taken three at a time, as follows. b. You can find the number of different poker hands by using the formula for the number of combinations of 52 elements taken five at a time, as follows.