Fundamentals of Engineering Exam Review of Combinations

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

He will start with defining a combination and move on to running through an example of how combinations are used in the statistics and probability portion of the exam.

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

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In the trenches working with those who have been away from the books for ages, from standardized testing for the same amount of time, and can’t fathom taking on this beast of an exam.

And every iteration in between.

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Transcription

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 Combinations.

When you need to count the number of groupings without regard to order then combinations is the way to go. Now in a previous tutorial, I went over permutations.

Permutations specifically count the number of ways a task can be arranged or ordered…that is the fundamental difference between the two.

A permutation is with regard to order and combinations is without regard to order.

So if you guys need a refresher on that you can go ahead and jump over to that tutorial once this one’s complete this one will be completely standalone so you can continue to watch and follow along.

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

A combination is an arrangement of our objects without

regard to order and without repetition

selected from n distinct objects is

called a combination so to put that in

into an equation a combination a

combination oops a combination of n

objects taken R at a time is equal to n

factorial divided by n minus R factorial

R factorial so the difference between

combinations and permutations again is

in combinations you are counting groups

order is not important and in

permutations you are counting different

ways to arrange items with regard to

order the N and R mean the same thing in

both permutation and combinations but

the formula differs and it differs by

this

our factorial in the denominator so

let’s define that one more time and

arrangement

of our objects

without regard and I will emphasize

without without regard to order and

without repetition

selected from selected from n distinct

objects

is called a combination

that’s it so without regard without

regard to order combinations with regard

to order permutation so let’s go through

an example to illustrate that let’s say

that we want to draw three marbles at

random from a bag containing three red

and five white marbles so we want to

take three marbles at random from a bag

that contains three red and five white

so this is the bag and the bag has eight

total marbles and we want to take three

so how many different draws or how many

different ways or how many different

ways can can three marbles come out of

this bag of eight this bag of eight

that’s broken up into three red and five

white this would be a combination

problem because the draw would be a

group of marbles without regard to order

it is like grabbing a handful of marbles

and just looking at them note that there

are no special conditions placed on the

marbles that we draw so this is

straightforward combination problem and

note that if we are putting these

marbles in any kind of order then we

would need to use permutations to solve

the problem but in this case order does

not matter so we’re going to use

combinations so let’s go ahead and

define N and R for a combination

equation n is the number of marbles we

can choose from in this bag once again

has eight total marbles so n is equal to

eight R is the number of marbles we will

be drawing at one time and we want to

choose three marbles so R is equal to

three and all we need to do is plug this

into our combination equation which is n

factorial divided by

n minus R factorial R factorial and

plugging that in n is equal to 8 and R

is equal to 3 and simplifying that we

got 8 factorial divided by 5 factorial

and 3 factorial so once again you can

simplify this further by recognizing

that 8 factorial is 8 times 7 times 6

times 5 times 4 times 3 times 2 all the

way down to 1 and then 5 factorial is 5

times 4 times 3 times 2 times 1 and 3

factorials 3 times 2 times 1 and notice

that we can just eliminate a couple of

these factors here because they just

divide out and our final answer is going

to be 8 times 7 times 6 divided by 3

times 2 times 1 and in this case there’s

it’s going to equal 56 and so there are

56 different draws that can come from

this bag of eight total marbles so let’s

move on with using this same bag of

marbles let’s move on to another example

to illustrate another scenario so let’s

say this time that we want to know how

many different draws would contain only

red marbles so remember there’s eight

total marbles and that three are red and

that five are white and out of these

eight marbles we’re still choosing three

at random

so once again this is a combination

problem because the draw would be a

group of the marbles and with without

regard to order and so on Part A we just

we determine how many different possible

draws there are which is 56 now we want

to know out of those 56 how many would

give us 3 red so our draw would look

like just three red marbles so let’s go

ahead and define N and R so in in this

case is n is the number of red marbles

we can choose from and in this case

there’s only three red marbles that are

in the bag so n is equal to 3 and R is

the number of red marbles that are drawn

at a time once again 3 R 3 are randomly

chosen but we want all three of those to

be red marbles so R is also going to

equal 3 so plugging that into our

combination equation n factorial divided

by n minus R factorial R factorial n is

equal to 3 factorial 3 minus 3 factorial

R factorial oops

R is equal to 3 this simplifies out the

3 factorial over zero factorial times 3

factorial and we can figure out that

that’s equal to 1 so out of the 56

possible draws only one draw would

contain 3 red marbles so let’s go one

more let’s go just a bit further to

complicated a little bit more but to

illustrate the use of combination in a

different scenario let’s use the same

bag of eight marbles so once again we

have eight marbles

three of which are red and five are

white so in this case how many different

drawers would contain one red marble and

two white marbles so we want one red

marble and we want two white marbles so

once again this is a combination problem

because this draw would be just a group

of marbles without regard to order and

like like we stated before in the first

part of this problem there’s 56 possible

drawers and of those drawers we want to

know how many of them would have one red

and two white so first we need to

determine n NR together that would make

up one draw we are going to have to use

the counting principle in this problem

to help us out note how one draw one

draw of three marbles is split into two

parts red and white we cannot combine

them together because we need a

particular number for each one so we

will figure out how many ways to get one

red marble and how many ways to get two

white marbles and use the counting

principle we will multiply these numbers

together to get our final possible

outcomes possible drawers so let’s start

out with one red so in in this case

would be the number of red marbles we

can choose from so in the bag there’s

three red marbles and our would be the

number of red marbles we are drawing at

a time and in this draw we want one red

marble so plugging that in to the

equations we get n is equal to one

and our sorry n is equal to 3 and R is

equal to 1 and so that gives us if we

calculate that out that gives us or

simplify that gives us 3 2 factorial

times 1 factorial

now before we calculate this out let’s

go ahead and just look at the white

scenario let’s say we want to whites so

in this case n is equal to 5 because

there’s five white marbles in the bag

and in this case we also want two of

those to the three that we choose to be

white so R is equal to two so plugging

that in to the equation we have n is

equal to five and we have R is equal to

two so to simplify that that’s five

factorial divided by 3 factorial times 2

factorial now to get the total number of

possible draws that would contain one

red and two white we use D we use the

counting rule and multiply these two two

possibilities together to get the final

value so that would give us 3 factorial

divided by 2 factorial times one

factorial for our red scenario right

here multiply that out by 5 factorial

divided by 3 factorial times 2 factorial

and so we can just use our calculator in

this case would probably be the most

simple way but you can simplify it by

you know simplifying each of these and

two simpler more simple terms so we got

3 over 1 multiplied by 5 times 4 divided

by 2 times 1 and that equals 30 so out

of the 56 possible draws that we

determine in the first part 30 of those

would contain one red and two white

marbles so that’s it that’s all for our Fundamentals of Engineering Exam Review of Combinations and what Combinations consist of I hope that

helped you guys out a little bit

refreshed it just a tiny bit if you guys

have any more questions don’t hesitate

to contact me over at engineering

training exam calm and be more than

happy to help you out in whatever way

you need so we’ll be talking too soon

too soon

—

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Combinations – Fundamentals of Engineering Exam Review

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