### Video Transcript

If π₯ plus π¦ π four equals 360 and
two π₯ plus π¦ factorial equals 5040, find π¦ππ₯.

The notation π¦ππ₯ means the
number of permutations of π₯ unique items taken from a collection of π¦ unique
items. Using factorials, we can calculate
πππ, where π and π are both nonnegative integers and π is greater than zero as
π factorial divided by π minus π factorial. The factorial of a positive integer
π is defined as π times π minus one factorial. Expanding this out, we see the π
factorial is the product of all of the positive integers between one and π
inclusive. Additionally, we define zero
factorial to be equal to one.

Looking back at our statement, we
see that two π₯ plus π¦ factorial equals 5,040. So this equation should uniquely
determine two π₯ plus π¦. Specifically, one times two times
three times four and so on until we multiply by two π₯ plus π¦ gives us 5,040. Instead of solving this equation
algebraically, letβs list the first several factorial numbers to make an educated
guess for the approximate value of two π₯ plus π¦.

One factorial is one. Two factorial is two. Three factorial is six. Four factorial is 24. And five factorial is 120. In fact, these first five factorial
numbers come up so often that itβs even worth memorizing them. Anyway, we can see that by the time
we reach five, weβre already at 120. And the rate of growth is only
increasing because each time weβre multiplying by larger and larger numbers. The difference between 120 and
5,040 is only a factor of about 50. And remember, π factorial is π
times π minus one factorial. This means that six factorial is
six times 120 and seven factorial is seven times six times 120 and so on.

But seven times eight is 56. So by the time we get to eight
factorial, weβll be multiplying 120 by a factor larger than 50. So eight factorial is bigger than
our target number, which means that two π₯ plus π¦ is either six or seven. Doing out the calculations, six
times 120 is 720 and seven times 720 is 5,040, exactly the number weβre looking
for. So since two π₯ plus π¦ factorial
is 5,040, which is seven factorial, it follows that two π₯ plus π¦ is seven.

This sort of trial and error
calculation is very common with factorials. But itβs worth mentioning that we
knew from the beginning that we wouldnβt have to go very far in our list before we
found what we were looking for. This is because factorial numbers
grow incredibly quickly. Note that we started with one
factorial equals one and by the time we got to seven, our number was greater than
5,000. Just three numbers later, we reach
10 factorial, which is larger than 3.5 million. Precisely because factorial numbers
grow so rapidly, we knew that we could quickly find 5,040 with a direct search.

Okay, so now we know that two π₯
plus π¦ is seven. But we need to know what π₯ and π¦
are if weβre going to calculate π¦ππ₯. So we need one more equation to
solve for these two unknowns. And we can find this equation using
the other equation that weβre given, π₯ plus π¦π four equals 360. Letβs clear out the list of
factorials to make room for our calculations.

The definition of πππ gives us
that 360 is equal to π₯ plus π¦ factorial over π₯ plus π¦ minus four factorial. Now weβre going to rewrite the
numerator. Specifically, since π factorial is
π times π minus one factorial, π factorialβs also π times π minus one times π
minus two factorial and so on. Weβre actually going to do this
four times so that weβll have an π₯ plus π¦ minus four factorial in both the
numerator and the denominator. So we get π₯ plus π¦ times π₯ plus
π¦ minus one times π₯ plus π¦ minus two times π₯ plus π¦ minus three times π₯ plus
π¦ minus four factorial all divided by π₯ plus π¦ minus four factorial.

To double-check that this is
correct, observe that from our definition this product is π₯ plus π¦ minus three
factorial. So this product is π₯ plus π¦ minus
two factorial. So this product is π₯ plus π¦ minus
one factorial. And finally, this product is π₯
plus π¦ factorial, which is exactly what it should be. Now π₯ plus π¦ minus four factorial
in the numerator divided by π₯ plus π¦ minus four factorial in the denominator is
just one. So we have 360 equals π₯ plus π¦
times π₯ plus π¦ minus one times π₯ plus π¦ minus two times π₯ plus π¦ minus
three.

Just like our earlier equation,
this equation would be quite hard to solve algebraically. Luckily, thereβs a nifty trick we
can use to get very close to the correct answer that relies on the fact that π₯ plus
π¦ is an integer and therefore these four factors are consecutive integers. Here are our four consecutive
integers drawn on a number line. And if we let π be π₯ plus π¦,
then these are exactly the four factors that we have that multiply to 360. The observation is that as long as
these integers are all greater than one, if we take the fourth root of their
product, this number will be on the number line somewhere between the middle two
integers.

In fact, in general, the πth root
of the product of π consecutive integers all greater than one will be very close to
the average value of those integers. So if we calculate the fourth root
of 360, we should get a number somewhere between π₯ plus π¦ minus one and π₯ plus π¦
minus two. But we can then use this fact to
come up with an approximate value for π₯ plus π¦ and then proceed by trial and
error. If we plug into a calculator, weβll
find that the fourth root of 360 is approximately 4.4. The exact number isnβt important,
but what is important is that this number is between four and five.

So since 4.4 is between four and
five, we expect that π₯ plus π¦ minus two is going to be four and π₯ plus π¦ minus
one is going to be five, which means π₯ plus π¦ should be six. And indeed, six times five times
four times three is exactly equal to 360. So π₯ plus π¦ equals six. Now we have two equations and two
unknowns, so letβs clear some space and solve for π₯ and π¦.

If we subtract π₯ plus π¦ equals
six from two π₯ plus π¦ equals seven, on the left-hand side two π₯ minus π₯ is π₯
and π¦ minus π¦ is zero and on the right-hand side seven minus six is one. So π₯ is one. If π₯ equals one and π₯ plus π¦
equals six, then π¦ is six minus one, which is five. Finally, to calculate π¦ππ₯, we
simply need to find five π one.

There are two ways to calculate
this. First, we can plug in five for π
and one for π in our formula for πππ. Or we can know the following very
useful identity. For all positive integers π, ππ
one is equal to π. This follows directly from the
definition of πππ and the definition of π factorial as π times π minus one
factorial. So without any calculation, we
already know that five π one is equal to five. And this is the answer weβre
looking for.