Original subtitles

-: Hello folks.

In this lecture we are going to look

at one of the most commonly used parts

of combinatorics, permutations.

Permutations represents the number

of different possible ways we can arrange a set of elements.

These elements can be digits, letters, objects,

or even people.

To clear any confusion, let's look at an example.

Imagine you haven't watched the latest Formula One race

but your friends spoiled who the three drivers

on the podium are;

Lewis, Max and Kimi.

A permutation of three, denoted P of three

would express the total number of different ways

these drivers could split the medals among one another.

Suppose Lewis won the race,

then we have two possible scenarios,

Max finished second and Kimi finished third.

Or, Kimi finished second and Max finished third.

Now suppose that Max won,

once again we have two possible outcomes,

but this time it is Lewis and Kimi

who have to split the silver and bronze medals.

Either Kimi got silver and Lewis got bronze,

or the other way around.

Not surprisingly, if Kimi won the race

we would have two more ways

the drivers can be arranged on the podium.

Either Max gets silver and Lewis gets bronze,

or Lewis gets silver and Max gets bronze.

In total this leaves us with six unique ways

these three drivers

can split the top three spots.

We call these six ways permutations

and we are going to show you how to compute the number

of permutations for a finite set of any size

and, in different situations.

Great. Now let's discuss the intuition behind computing

the total number of permutations

for a set of n many elements.

We start filling out the positions one by one.

The order in which we fill them out is completely up to us.

For convenience, we usually start with the first slot,

which represents the race winner in our example.

Since anybody out of the n many drivers

in the set could have won the race,

we have n different possible winners.

After that, we have n minus one possible drivers left

and any one of those can finish second.

Regardless of which out of the n elements we chose to

take the first slot, we have n minus one

many possibilities for the second slot.

Similarly, we would have n minus two possible outcomes

for who finishes third and so on.

Generally, the further down the ranking we go

the more options we exhaust, and the more options we exhaust

the fewer options we have left.

This trend will continue until we get to the last element,

for which we will only have a single option available.

Therefore, mathematically, the number of

permutations would equal the product of n, n minus one,

n minus two, and so on, until one.

We denote this product as n factorial, and

in the next video we are going to discuss

some of its main properties and how these properties

relate to permutations.

See you there and thanks for watching.