05.04.2019: One Lesson of Math - Permutations and Combinations, 2/5: Permutations Involving Differen


Today's soundtrack is The Red Chord: Clients, a technical metalcore album.


May the Fourth be with you, and also with you.


This morning, I'm learning about permutations, which are ways that we can rearrange the items in a series.


This concept was touched on in yesterday's sandwich example. The series B E D has the following permutations: BED, BDE, EBD, EDB, DEB, and DBE. We can break it down like this: We have three choices for our first letter, two choices for our second letter, and a single choice for the third letter, giving us the equation 3⋅2⋅1=6 permutations.

"Surely," you say to yourself, "there must be some more efficient way to write all this out?"


Yes, dear reader, there is, and it has a name: Factorial Notation.


In factorial notation, we use n! to indicate multiplication of all the integers between 1 and n, inclusive. So 5! = 5⋅4⋅3⋅2⋅1; 3! = 3⋅2⋅1, and so on.


Hidden in the [math] section of my TI-83 Plus is the factorial function, which is useful when dealing with large factorials (such as 21!). To use it, we first type the number that we want to factorialize, hit [math], go to the PRB menu, and choose "!". Then we can hit "enter" to calculate the factorial of the desired number.


What if we have lots of individual items that we want to arrange in a specific way? For example, I want to buy a pizza with two toppings. There are five topping options available to me. Using the fundamental counting principle, I could determine that I have five choices for my first topping and four choices for my second topping. 5⋅4=20. What if I want to write this out in factorial notation? 5! = 5⋅4⋅3⋅2⋅1, but we need to cancel out 3⋅2⋅1. 3! = 3⋅2⋅1. So if we write 5!/3!, we will get our answer: 20 permutations. The general equation for this concept - taking r of n distinct objects at a time - is n! / (n-r)!. This equation equals ₙPr , which means n permute r. So we could write out our pizza choices from the above example as ₅P₂, because we have 5 distinct objects permuted 2 at a time.


As with the factorial notation, ₙPr can be accessed in the PRB menu of the [math] function section on the TI-83 Plus. To use it, first type the number that equals n, then choose the function, and finally type the number for r and then hit "enter" to get the result.


When working with permutations in the form nPr, we must keep several restrictions in mind:

  • 0! = 1

  • Any factorial divided by 0! will equal 1

  • Any integer divided by 0! will equal itself

  • n ≥ 0

  • n r

  • r ≥ 0


If we are asked to write a factorial with neither factorial notation nor the nPr notation when r is known, we can break it down like this:


First, look at the lowest value of r. That is how low we will take n. So nP3 = [n ⋅ (n-1) ⋅ (n-2) ⋅ (n-3)] / [(n-3)]. The two (n-3)s cancel each other out, and we are left with n⋅ (n-1) ⋅ (n-2).


But what if we're dealing with binomials? For example, how would we work out (n + 4)! / (n-2)!? The same principle applies. Remembering how factorials work, we can rewrite (n + 4)! as (n+4) ⋅ (n+3) ⋅ (n+2) ⋅ (n+1) ⋅ n ⋅ (n-1) ⋅ (n-2). We would rewrite the denominator as (n-2). The two (n-2)s cancel each other out; our solution is (n + 4)! as (n+4) ⋅ (n+3) ⋅ (n+2) ⋅ (n+1) ⋅ n ⋅ (n-1).


If we are asked to solve for the unknown variable of a factorial, we can do so algebraically. See my examples below. Please disregard the eraser marks in section b); I had a brain fart and forgot how multiplication works.



That's it for today; next time, we'll be learning about permutations involving identical objects.

#TheRedChord