1. Advanced Math Topics 4.4 Permutations

2. You have six younger brothers, George 1, George 2, George 3, George 4, George 5, and George 6. They all want your help with their math homework. You only have time to help two brothers, one before dinner and one after. How many ways can this be done? First of all, does order matter? Yes, helping 1 before dinner then 2 after is different than helping 2 before dinner and 1 after. Note: We will refer to George 1 as 1, George 2 as 2,…etc. Possibilities: 1,2 1,3 1,4 1,5 1,6 2,1 2,3 2,4 2,5 2,6 3,1 3,2 3,4 3,5 3,6 4,1 4,2 4,3 4,5 4,6 5,1 5,2 5,3 5,4 5,6 There are 30 possibilities!! There are 6 people and you are picking 2 (6 pick 2). 6,1 6,2 6,3 6,4 6,5 You could find the same answer by calculating… 6! (6 – 2)! 6x5x4x3x2x1 4x3x2x1 6! 4! == total total – picks == 30

3. Permutation:An arrangement of distinct objects in a Particular order Permutations are Particular Picks. Factorial examples:7! =7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040 4! =4 x 3 x 2 x 1 = 24 1! =1 0! =1 This makes formulas involving 0! meaningful and will be explained further. The number of permutations of “n” things picked “r” at a time is… n P r = n! (n – r)! Why is this true and why does 0! = 1? The next slides can answer these questions. (order matters)

4. Al, Bob, Carey, Dee, and Eric have 5 front row seats at a concert. They each can sit in any of the 5 seats. How many seating possibilities are there? A person cannot sit in two seats, thus it is implied that there are no repetitions and order does matter. 5 x120 possibilities 4 x3 x2 x1 = Remember this question?.... There are 5 people being picked in a particular order 5 at a time. 5 pick 5 = 5 P 5 = 5! (5 – 5)! = 5! 0! = 5! 1 = 120 This can help explain why 0! = 1. n P n = n!

5. Al, Bob, Carey, Dee, and Eric are the only entries in a drawing for two front-row tickets to a concert. How many seating possibilities are there if two of them are picked? How does this change the solving process? 5 x20 possibilities 4= What if we changed the question slightly?.... There are 5 people being picked in a particular order 2 at a time. 5 pick 2 = 5 P 2 = 5! (5 – 2)! = 5! 3! = 5x4x3x2x1 3x2x1 = 20 There are two ways to solve this and it may help explain the permutation formula.

6. You are putting 2 identical clown dolls and 4 identical leprechaun dolls on the shelf. In how many different ways can this be done? Does this work? 6 x5 x4 x3 x2 x1 =6! =720 ways This means that in the first spot you have 6 choices, then you have 5 choices, etc. But some of the dolls are identical so this number is inflated! Switch the 1 st and 3 rd leprechaun to get this… They are the same arrangement but the solution above counts both of them. Thus, we need a new way to solve this to get fewer possibilities than 720. The number of permutations of “n” things of which “p” are alike, “q” are alike, or “r” are alike, and so on, is… n! p!q!r! where p + q + r…. = n. Using this formula: 6! 4!2! = 6x5x4x3x2x1 (4x3x2x1)(2x1) = 15

7. HW P. 205 #1-12 Let’s do #3, #10, and #8a together