Question Video: Evaluating Permutations and Factorials to Find the Values of Unknowns Then Evaluating Their Values in a Permutation Mathematics

If (π‘₯ + 𝑦)𝑃4 = 360 and (2π‘₯ + 𝑦)! = 5040, find 𝑦𝑃π‘₯.


Video Transcript

If π‘₯ plus 𝑦 𝑃 four equals 360 and two π‘₯ plus 𝑦 factorial equals 5040, find 𝑦𝑃π‘₯.

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

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

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 it’s even worth memorizing them. Anyway, we can see that by the time we reach five, we’re already at 120. And the rate of growth is only increasing because each time we’re multiplying by larger and larger numbers. The difference between 120 and 5,040 is only a factor of about 50. And remember, 𝑛 factorial is 𝑛 times 𝑛 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, we’ll be multiplying 120 by a factor larger than 50. So eight factorial is bigger than our target number, which means that two π‘₯ plus 𝑦 is either six or seven. Doing out the calculations, six times 120 is 720 and seven times 720 is 5,040, exactly the number we’re looking for. So since two π‘₯ plus 𝑦 factorial is 5,040, which is seven factorial, it follows that two π‘₯ plus 𝑦 is seven.

This sort of trial and error calculation is very common with factorials. But it’s worth mentioning that we knew from the beginning that we wouldn’t 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 𝑦 is seven. But we need to know what π‘₯ and 𝑦 are if we’re going to calculate 𝑦𝑃π‘₯. So we need one more equation to solve for these two unknowns. And we can find this equation using the other equation that we’re given, π‘₯ plus 𝑦𝑃 four equals 360. Let’s clear out the list of factorials to make room for our calculations.

The definition of π‘›π‘ƒπ‘Ÿ gives us that 360 is equal to π‘₯ plus 𝑦 factorial over π‘₯ plus 𝑦 minus four factorial. Now we’re going to rewrite the numerator. Specifically, since 𝑛 factorial is 𝑛 times 𝑛 minus one factorial, 𝑛 factorial’s also 𝑛 times 𝑛 minus one times 𝑛 minus two factorial and so on. We’re actually going to do this four times so that we’ll have an π‘₯ plus 𝑦 minus four factorial in both the numerator and the denominator. So we get π‘₯ plus 𝑦 times π‘₯ plus 𝑦 minus one times π‘₯ plus 𝑦 minus two times π‘₯ plus 𝑦 minus three times π‘₯ plus 𝑦 minus four factorial all divided by π‘₯ plus 𝑦 minus four factorial.

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

Just like our earlier equation, this equation would be quite hard to solve algebraically. Luckily, there’s a nifty trick we can use to get very close to the correct answer that relies on the fact that π‘₯ plus 𝑦 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 𝑦, 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 𝑛th root of the product of 𝑛 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 𝑦 minus one and π‘₯ plus 𝑦 minus two. But we can then use this fact to come up with an approximate value for π‘₯ plus 𝑦 and then proceed by trial and error. If we plug into a calculator, we’ll find that the fourth root of 360 is approximately 4.4. The exact number isn’t 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 𝑦 minus two is going to be four and π‘₯ plus 𝑦 minus one is going to be five, which means π‘₯ plus 𝑦 should be six. And indeed, six times five times four times three is exactly equal to 360. So π‘₯ plus 𝑦 equals six. Now we have two equations and two unknowns, so let’s clear some space and solve for π‘₯ and 𝑦.

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

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

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