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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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