All language subtitles for 05 - Calculate permutations with duplication

- [Instructor] In the previous movie,

I described how to work with permutations,

that is sets where the order

of values matters, without duplication.

In this movie, I will show you how to calculate the number

of possible permutations when duplication is allowed.

As before, let's assume that we have eight speakers

who are available to speak at a company event

and we want to choose three.

However, it is possible that we can have

the same person go twice.

So in other words, duplication is allowed.

I do want to emphasize that order still matters.

So having the same person go first and third

is considered a different permutation

than having a person go first and second.

So once again, order still matters.

So our question here is how do we calculate the number

of unique orders that are possible when choosing three

of eight people when repetition is allowed?

To demonstrate the process, I will switch over

to our Excel sample file.

I'm now in Excel and I've opened our sample workbook,

which is 06_05, Permutations With Duplication.

You can find that in the chapter six folder

of the exercise files collection.

I have defined our scenario with eight available speakers

and three available speaking slots.

And for comparison, I left the value

from the previous movie, Permutations Without Duplication,

in cell B6 and remember that that is 336 possible orders.

Now let's see how many we get if we have permutations

with duplication allowed.

So I'll click back in cell B8, type an equal sign.

And the function we will use is permutation A.

That is the second one available here

after I type in P-E-R-M-U.

And permutation A returns the number of permutations

for a given number of objects when repetition is allowed.

So I'll press Tab and the number of possible items is B3.

That's the collection we are choosing from.

Then the comma and the number chosen.

In this case, the number of available

speaking slots is in B4.

Right parentheses and Enter. And we get 512.

So significantly more.

However, it is limited by the number of open speaking slots.

In the previous movie, I asked what would happen

if we had 10 available speakers

and five open speaking slots.

So I will change the value in B3 to 10.

And we'll go from 336 and 512 to 720 and 1,000.

So quite a few.

And if I increase the number of open speaking slots,

we go from 720 and 1,000 to 30,240 and 100,000.

Mathematicians call this combinatorial explosion.

And I think you can see why because the numbers

get very large, very fast.