All language subtitles for 004 Solving Variations with Repetition_en

-: Welcome back.

So far, we explained what permutations are

and how to compute them.

In this lecture, we are going to focus

on a similar but not too similar concept.

Variations.

Variations, express the total number of ways we can pick

and arrange some elements of a given set.

For example, imagine you went on vacation

and forgot the code for the combination lock

on your carry on.

Luckily for you, the lock requires a two

letter code using only the letters A, B, and C to unlock it.

We can approach the problem as we did before

and explore the different positions in a specific order.

Let us start with the first letter.

We have three different options A, B, or C.

Suppose we chose A and move on to the next letter.

Since we can repeat values

we once again have the same three options

for the second letter A, B, or C.

This indicates that there are three different variations,

if we decide to start with A.

Now, if we put B in the first position,

again we would have three options

for what we choose for the second letter.

In general, regardless of which one

of the three letters we decide to start with

we are going to have three different options

for the second letter.

Therefore, the total number of variations we can get

is three times three equals nine.

The formula we use to calculate variations

with repetition is the following.

V bar of N and P equals N to the power of P

where N is the total number of elements we have available

and P is the number of positions we need to fill.

The way we interpret this notion is

the number of variations with repetition when picking P

Many elements out of N elements is equal

to N to the power of P.

If we apply this to the combination lock example

we would write V bar of three and two equals three

to the power of two, which is equal to nine.

We interpret this as there are nine different variations

of two letter pass codes consisting of A, B, or C only.

What happens if the law could use any of the 26 letters?

we would have 26 to the power of two

which is 676 different variations.

All right, now you know how to deal

with variations with repeating values.

Make sure to complete the exercises

after this lecture to test your knowledge

and reinforce what you've learned.

In the next lecture

we are going to briefly explain what variations

without repetition are and how to compute them.

I hope to see you there and thanks for watching.