Counting Principals, Permutations, and Combinations

A counting problem asks “how many ways” some event can occur. Ex. 1: How many three-letter codes are there using letters A, B, C, and D if no letter can be repeated? One way to solve is to list all possibilities. ABC ABD ACB ACD ADB ADC BAC BAD BCA BCD BDA BDC CAB CAD CBA CBD CDA CDB DAB DAC DBA DBC DCA DCB

Ex. 2: An experimental psychologist uses a sequence of two food rewards in an experiment regarding animal behavior. These two rewards are of three different varieties. How many different sequences of rewards are there if each variety can be used only once in each sequence? Next slide

Count the number of branches to determine the number of ways. Another way to solve is a factor tree where the number of end branches is your answer. Count the number of branches to determine the number of ways. There are six different ways to give the rewards.

Fundamental Counting Principle Suppose that a certain procedure P can be broken into n successive ordered stages, S1, S2, . . . Sn, and suppose that S1 can occur in r1 ways. S2 can occur in r2 ways. Sn can occur in rn ways. Then the number of ways P can occur is

Ex. 2 Revisited: An experimental psychologist uses a sequence of two food rewards in an experiment regarding animal behavior. These two rewards are of three different varieties. How many different sequences of rewards are there if each variety can be used only once in each sequence? Using the fundamental counting principle: 3 2 X 1st reward 2nd reward

Permutations An r-permutation of a set of n elements is an ordered selection of r elements from the set of n elements ! means factorial Ex. 3! = 3∙2∙1 0! = 1

Note: The order does matter Ex. 1 Revisited: How many three-letter codes are there using letters A, B, C, and D if no letter can be repeated? Note: The order does matter

The number of combinations of n elements taken r at a time is Order does NOT matter! Where n & r are nonnegative integers & r < n

Ex. 3: How many committees of three can be selected from four people? Use A, B, C, and D to represent the people Note: Does the order matter?

Ex. 4: How many ways can the 4 call letters of a radio station be arranged if the first letter must be W or K and no letters repeat?

Ex. 5: In how many ways can a class of 25 students elect a president, vice-president, and secretary if no student can hold more than one office?

Ex. 6: How many five-card hands are possible from a standard deck of cards?

Ex. 7: In how many ways can 9 horses place 1st, 2nd, or 3rd in a race?

Ex. 8: Suppose there are 15 girls and 18 boys in a class Ex. 8: Suppose there are 15 girls and 18 boys in a class. In how many ways can 2 girls and 2 boys be selected for a group project? 15C2 × 18C2 = 16,065

It’s time to combine & expand From a group of 7 men and 6 women, five people are to be selected to form a committee so that at least 3 men on the committee. In how many ways can it be done? What are the possible scenarios 3 men and 2 women, 4 men and 1 woman, or 5 men (7C3 x 6C2) + (7C4 x 6C1) + (7C5) - but why do we add here? (525 + 210 + 21) 756

MORE EXAMPLES In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together? The word 'LEADING' has 7 different letters. But, when we pair all the vowels together (EAI) they count as 1 letter So there are 5 letters → 5! = 120 combinations But the vowels can also be rearranged, so → 3! = 6 When we combine these two we get 120 x 6 = 720 possible solutions

Continued In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions? There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants. Let us mark these positions as under: (1) (2) (3) (4) (5) (6) Now, 3 vowels can be placed at any of the three places marked 1, 3, 5. Number of ways of arranging the vowels = 3P3 = 3! = 6. Also, the 3 consonants can be arranged at the remaining 3 positions. Number of ways of these arrangements = 3P3 = 3! = 6. Total number of ways = (6 x 6) = 36.