All language subtitles for 005 Solving Variations without Repetition_en

Narrator: Welcome back, everyone.

We've already covered how to calculate variations

for events with repetition.

In this lecture, we are going to focus on events

where we apply variations without repetition.

Okay, let's begin.

Imagine you are a track and field coach

and need to choose which four members of your team

run the relay and in what order.

The team consists of five people, Tom, Eric, David, Kevin,

and Josh, and you must decide who starts, who anchors,

and who runs in between.

Okay.

We have five members in the team,

which means five different scenarios

for who we want to start.

Let's say we know that David is the best guy for the job.

That decision leaves us with only four options

for who gets the second position,

namely Tom, Eric, Kevin, or Josh.

Suppose we chose Josh to run after David.

That leaves us with only three options

for who runs third, Tom, Eric, or Kevin.

Now, if we pick Kevin, then we know one

of either Tom or Eric finishes the race.

Not surprisingly, if we chose Eric to run third instead,

we once again have two choices

for who runs last, Tom or Kevin.

Finally, if we choose Tom to run third,

we would have two options available.

Eric and Kevin.

This means we can have six different variations

for who runs the last two positions

if we have chosen David and Josh to run first

and second respectively.

Using a similar logic, if we pick somebody different

than Josh to run second,

we would still have six possible variations

for who runs third and fourth.

In fact, regardless of who out of the four members

of the team we chose to run second,

we would always have six options

for who fills out the remaining spots on the team.

That suggests, there are four times six or 24 different ways

to arrange the three remaining positions

if we knew David starts.

What if we aren't sure that David

is the best runner to start?

Well, if somebody out of the remaining four people started,

we would still have 24 ways of filling out

the remaining spots on the team.

What happens is, the further down the order we go,

the fewer options we are left with,

since nobody is permitted to run multiple legs.

This is what variations without repetition is about.

We cannot use the same element

or in this case, person twice.

In terms of numbers,

we have five times four times three times two.

This makes 120 different options of how to arrange

our team for the competition.

That wasn't too bad, right?

It's time to introduce you to the formula

for calculating variations without repetition.

The number of variations without repetition

when arranging p elements out of a total of n

is equal to n factorial over n minus p factorial.

Applying this formula to our example,

the total number of variations

without repetition would be equal to five factorial

over one factorial.

Using the properties we introduced

in the bonus lecture about factorials.

This is equal to two times three

times four times five, which equals 120.

All right, let's wrap up here.

As usual, please go through the homework exercises

and try answering all the quiz questions.

Thanks for watching.