In this episode of Engineer In Training Exam TV, Justin walks you through a Fundamentals of Engineering Exam Review of Permutations.

We will begin by defining a factorial, move on to defining a Permutation, and then run through an example of how permutations are calculated in the statistics and probability portion of the exam.

This Fundamentals of Engineering Exam Review of Permutations is part of the global subject Probability and Statistics.

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Hey what’s going on everyone, it’s Justin Dickmeyer from EngineerInTrainingExam.com.

In today’s video we are going to present a Fundamentals of Engineering Exam Review of Permutations.

Permutations are an offshoot of the fundamental counting principle which I covered in a previous review. Permutations are used to specifically count the number of ways a task can be arranged or ordered.

So let’s start off our Fundamentals of Engineering Exam Review of Permutations by defining exactly factorials are.

Factorial written as in or any

number exclamation point that’s n

factorial and n factorial is the product

of whatever n is n times n minus 1 times

n minus 2 all the way down to 1 so let’s

say you want to get 7 factorial it would

first be written as 7 factorial and say

you want the product or what that equals

the answer to that would be 7 times 6

times 5 times 4 all the way down to 1

and so the answer to that if you just

put that into your calculators real

quick would be 5040 so real quick before

we go on factorials are pretty basic and

I know you guys all know them but it

just wanted to touch on those real quick

but one special definition of factorial

is we need to know that 0 factorial is

always equal to 1 so if N in feature

problems we see zero factorial it’s not

that we’re dividing by 0 that actually

is equal to 1 so moving on what is a

permutation now a permutation is an

order of arrangements of our objects

without repetition selected from n

distinct objects is called a permutation

of n objects

can are at a time and that is denoted as

in P R is equal to n factorial divided

by n minus R factorial so let’s define

that again and order and that’s

important so make sure you underline

that an order of arrangements so I

running out of room here an order of

arrangements of our objects selected or

taken from n distinct objects is called

a permutation of n objects taken R at a

time so that might look a little

confusing butting in equation form you

can just note that the permutation of n

objects taken R at a time is equal to n

factorial divided by n minus R factorial

so to simplify my complicated definition

here that might have confused you guys

sorry about that but to simplify when

you need to count the number of ways you

can arrange items where order I bring

that up again where order is important

then you can use permutations to count

so once again when you need to count the

number of ways you can arrange items

where order is important then you can

use permutation to count so let’s look

at it let’s illustrate this at using in

an example here so let’s say that we

have 8 pictures and we want to hang all

eight of those pictures on to the wall

so the question would be in how many

ways

can eight pictures be hung on the wall

so since we are arranging these pictures

on the wall that means order is

important and because order is important

we can use permutations to count how

many ways so first we need to we need to

define the number of objects or pictures

and so in like I said is the pictures in

this in this example so since there’s

eight pictures in is going to equal

eight now are on the other hand is the

number of pictures we are going to use

at a time so once again we want to hang

all eight pictures on the wall so all

eight pictures are going to be used at

the same time so R is going to equal 8

so taking our permutations equation that

we defined before we got n factorial

divided by n minus R factorial and all

we need to do is plug in our our numbers

or our values for this specific problem

so we have eight objects so n is going

to equal 8 and we want to hang all eight

of them on the wall at the same time so

R equals 8 so we can simplify that as 8

over 0 factorial and once again I said

zero vectorial is equal to 1 so we can

go ahead and just simplify this as 8

times 7 times 6 times 5 times 4 all the

way down to 1

and when we multiply that all out we

find that there it’s 40,320 so that’s a

very large number and that tells us that

there’s four 40,320 ways that eight

pictures can be hung on the wall all at

the same time so let’s move on to

another example let’s say that we have

an engineering group and that

engineering group has 20 members and out

of those 20 members that engineering

group needs to select a president it

needs to select a vice president and it

needs to select a speaker and assuming

that the same person cannot hold that to

position two positions at the same time

assuming that a member can hold only one

position at a time that’s our assumption

so one member can’t be vice president

and president so we can go move forward

and define our values in an AR and once

again since we are choosing positions

this is a way to rank members and rank

equals order so order is important in

this problem once again so we can use

the equations of permutations to

determine the number of ways that three

members can be chosen from the twenty

within the group so we first need to

define n so n is the number of members

we can choose from and in this case we

have 20 members of those 20 members we

want to choose 3 for this the positions

the president the vice president and the

speaker so since we’re choosing 3 at a

time

r is going to equal 3 so once again all

we need to do is take our standard

permutation equation

we got NP R is equal to n factorial

divided by n minus R factorial and we

know that n is equal to 20 and we know

that R is equal to 3 so to simplify that

we got 20 factorial over 17 factorial

now let’s just say you know most of us

are going to be using the factorial key

on our calculators to determine the what

what these equations are equally and out

to be but let’s just say you you don’t

have that just for conversations sake

here you can simplify this equation so

you don’t have to write it out as 20

times 19 times 18 times 17 etc and since

I already wrote it out like that let me

show you how you can simplify it so

let’s expand on 20 factorial that’s 20

times 19 times 18 times 17 times 16 all

the way down to 117 factorial 17 times

16 times 15 times all the way down to 1

now all we need to need to notice here

is that we can cancel out every value

from the denominator that’s on the

numerator and starting at 17 all the way

down so our answer here will really be

20 times 19 times 18 and in this case

that equals 6 that 68 40 so running the

calculations this tells us that there

are six thousand eight hundred and forty

different ways to select three members

of the 20 member group for the three

positions so that’s it guys that’s all

there is to permutations I know it might

sound a bit complicated just the word in

itself but it’s really all not not that

big of a deal once again were concerns

with order in these problems so let me

just tell you once again that the

definition is an order of arrangements

of

are objects without repetition selected

from n distinct objects is called a

permutation so that’s it if you guys

have any more questions or need any more

guidance for preparing for the

engineering training exam don’t hesitate

to visit my site engineer and training

exam comm and I look forward to meeting

you guys and helping you in any way that

I’m able so for now take care and look

forward to talking to you soon

—

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