MATH 1107 Elementary Statistics Lecture 7 Combinations and Permutations

You have lost your debit card. Assuming that your signature cannot be forged, and the thief tries to discover your PIN, how many combinations will he have to try? Most PIN-based cards, have 4 digit PINs. Assuming that each number (0-9) can be used in each position, there are 10*10*10*10 or 10 4 or 10,000 different ways. This is an example of the fundamental counting rule. The answer is M n, where M is the number of possibilities of an event and n is the number of events.

Combinations and Permutations How many different ways can a student answer all the questions on a true-false test with 20 questions? The answer is 2 20 or 1,048,576 different ways. In other words, the probability of getting the right answer at random is 1/1,048,576.

Combinations and Permutations You are having a sit-down dinner with 10 people. How many different ways can you arrange the seating at a single table? This is called a permutation problem – The answer is (n)*(n-1)*(n-2)*(n-3)…or n!. In this problem, the answer is 10! or 10*9*8*7*6*5*4*3*2*1 or 3,628,800.

Combinations and Permutations Now, what if you only have 8 chairs? How many different ways can you seat 10 people with 8 chairs (2 people will have to sit on the floor or stand)? This too is a permutation problem – with n distinct objects taken r at a time. In this instance, we have 10 objects (people) taken 8 at a time: n P r = n!/(n-r)!

Combinations and Permutations So, we have 10 people (n) taken 8 at a time ®… 10!/(10-8)! = 3,628,800/2 or 1,814,400

Combinations and Permutations The Liberals for Bush club as 24 members. If four names are selected from a hat to represent the four officers of the club, how many different ways can this be done? In this instance, we have 24 objects or people (n) taken 4 at a time (r) : n!/(n-r)! = 24!/(24-4)! or 24!/20! = 24*23*22*21 = 255,024.

Combinations and Permutations How many different permutations are there of the letters in the word “book”? This question can be approached two ways… If we determine that the first “o” and the second “o” are unique letters, then we treat the problem as a factorial problem – n! or 4*3*2*1 = 24.

Combinations and Permutations However, if we determine that the two “o”s are not different (order does not matter), then the approach is slightly different. There will be fewer possibilities, if we do not discriminate between the “o”s – n!/(n 1 !*n 2 !*n 3 !) Specifically, in this case, we have 4 letters (n), 1 b (n 1 ), 1 k (n 2 ) and 2 o (n 3 )… 4!/(1!*1!*2!) = 24/2 = 12.

Combinations and Permutations Notice that in all of the previous problems, different orderings counted (BOOK is different from KOOB). When different orderings of the same events are counted separately, we have a permutation. Alternatively, when orderings of the same events are not counted separately, we have a combination. In a combination, BOOK is considered to be the same as KOOB.

Combinations and Permutations7 In how many different ways can a person gathering data for a market research firm select 3 of 20 households in an apt complex for interviewing? In this instance, we don’t care what order the households are selected (1,2,3 is the same as 3,2,1). So the formula is slightly different: n C r = n!/r!(n-r)!

So, we have 20 households (n) taken 3 at a time (r) … 20!/3!(20-3)! = 1,140 Combinations and Permutations Remember that there will ALWAYS be fewer possible combinations than permutations!