Advanced Math Topics 4 4 Permutations You have

Advanced Math Topics 4. 4 Permutations

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? Note: We will refer to George 1 as 1, George 2 as 2, …etc. First of all, does order matter? Yes, helping 1 before dinner then 2 after is different than helping 2 before dinner and 1 after. There are 30 possibilities!! Possibilities: 1, 2 3, 1 5, 1 There are 6 people and you are picking 2 (6 pick 2). 1, 3 3, 2 5, 2 1, 4 3, 4 5, 3 You could find the same answer by calculating… 1, 5 3, 5 5, 4 total 6! 6! 6 x 5 x 4 x 3 x 2 x 1 1, 6 3, 6 5, 6 = = 30 2, 1 4, 1 6, 1 total – picks (6 – 2)! 4! 4 x 3 x 2 x 1 2, 3 4, 2 6, 2 2, 4 4, 3 6, 3 2, 5 4, 5 6, 4 2, 6 4, 6 6, 5

Permutation: An arrangement of distinct objects in a Particular order Permutations are Particular Picks. (order matters) Factorial examples: 7! = 4! = 1! = 0! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040 4 x 3 x 2 x 1 = 24 1 1 This makes formulas involving 0! meaningful and will be explained further. (order matters) 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.

Remember this question? . . 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. 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! This can help explain why 0! = 1. 5! 1 = n. P n = 5 x 4 x 3 x 2 x 1 = 120 possibilities 120 n!

What if we changed the question slightly? . . 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? There are two ways to solve this and it may help explain the permutation formula. 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! 5 x 4 5 x 4 x 3 x 2 x 1 = 20 possibilities

You are putting 2 identical clown dolls and 4 identical leprechaun dolls on the shelf. In how many different ways can this be done? 6 x 5 x 4 x 3 x 2 x 1 = 6! = 720 ways Does this work? 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! = 6 x 5 x 4 x 3 x 2 x 1 (4 x 3 x 2 x 1)(2 x 1) = 15

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