Question Video: Evaluating Permutations to Find Unknowns Mathematics

subscript (𝑛 βˆ’ π‘š)𝑃₃ = 35,904 and 𝑛 + π‘š = 60. Find π‘š and 𝑛.

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Video Transcript

𝑛 minus π‘šπ‘ƒ three equals 35,904 and 𝑛 plus π‘š equals 60. Find π‘š and 𝑛.

We’re looking for the two unknowns π‘š and 𝑛. So we’ll need two equations. We are given two equations in the statement. Of these, 𝑛 plus π‘š equals 60 is a straightforward relationship between 𝑛 and π‘š. So our first step will be to manipulate 𝑛 minus π‘šπ‘ƒ three equals 35,904 into something more manageable. If π‘˜ and π‘Ÿ are nonnegative integers with π‘˜ greater than or equal to one, then the notation π‘˜Pπ‘Ÿ means the number of permutations of π‘Ÿ unique objects taken from a collection of π‘˜ unique objects and can be calculated as π‘˜ factorial divided by π‘˜ minus π‘Ÿ factorial.

The factorial of a positive integer π‘˜ is the product of all of the integers from one to π‘˜ inclusive. And so it obeys the recursive relationship π‘˜ factorial is equal to π‘˜ times π‘˜ minus one factorial. Additionally, we define zero factorial as equal to one. Observe that we can use our expression for the factorial of a number to expand our expression for π‘˜Pπ‘Ÿ. Specifically, we can rewrite π‘˜Pπ‘Ÿ as π‘˜ times π‘˜ minus one factorial divided by π‘˜ minus π‘Ÿ factorial. But we can now expand this further because using this definition, π‘˜ minus one factorial is π‘˜ minus one times π‘˜ minus two factorial.

Using the definition again, we can replace π‘˜ minus two factorial with π‘˜ minus two times π‘˜ minus three factorial. If we repeat this process a total of π‘Ÿ times, we’ll see that the numerator is π‘˜ times π‘˜ minus one times π‘˜ minus two times π‘˜ minus three et cetera all the way to π‘˜ minus π‘Ÿ plus one times π‘˜ minus π‘Ÿ factorial. But the factor of π‘˜ minus π‘Ÿ factorial in the numerator divided by the factor of π‘˜ minus π‘Ÿ factorial in the denominator is just one. So this whole thing reduces to the product of the π‘Ÿ consecutive integers, π‘˜ times π‘˜ minus one times π‘˜ minus two et cetera all the way to π‘˜ minus π‘Ÿ plus one.

Now, usually we use the factorial form because it is far more compact. However, when we try to solve equations, as a general rule, we do need to expand out these factorials. So knowing that we can directly expand π‘˜Pπ‘Ÿ as these π‘Ÿ terms is very useful. In particular, the equation that we’re given has π‘Ÿ equals three. So we only need the three terms 𝑛 minus π‘š, 𝑛 minus π‘š minus one, and 𝑛 minus π‘š minus two. So now we have to solve 𝑛 minus π‘š times 𝑛 minus π‘š minus one times 𝑛 minus π‘š minus two equals 35,904. It’s worth mentioning that it would’ve been much more difficult to arrive at this equation had we had to go through the full process of evaluating the factorials.

To use this equation to help us find π‘š and 𝑛, we’ll try to solve for the term 𝑛 minus π‘š using a modified trial and error approach. Specifically, we know that the three terms on the left-hand side are consecutive integers all greater than one. Now we could just select random values for 𝑛 minus π‘š and plug the product into a calculator until we come up with the correct answer. In fact, if we’re clever about what values we choose, we can guarantee that we’ll find the correct answer in a fairly short time.

However, there is a much more direct way to approximate the correct answer. Our three integers multiply to give some product. But this means that if we take the cubed root of the product, this will be very close to the average value of the three integers. In fact, this works no matter how many integers we’re multiplying. For example, the 10th root of the product of 10 consecutive integers is very close to the average value of those 10 integers. If we plug the cubed root of 35,904 into a calculator, we find that it is very close to 32.99. 32.99 is very close to the integer 33. But this means that 33 is a very good guess for 𝑛 minus π‘š minus one, the average value of our three integers.

Note that 𝑛 minus π‘š minus one is also the middle number consecutively because for a group of consecutive integers, the median and the arithmetic mean are the same. Anyway, if 𝑛 minus π‘š minus one is 33, then 𝑛 minus π‘š minus two is 32 and 𝑛 minus π‘š is 34. And in fact, if we multiply 34 times 33 times 32, we find that this product is exactly equal to 35,904. So 𝑛 minus π‘š is 34. Now we have exactly what we’re looking for. 𝑛 plus π‘š equals 60 is a simple linear relationship between 𝑛 and π‘š and so is 𝑛 minus π‘š equals 34. So now we just need to use these two equations to find the two unknowns.

If we start with 𝑛 plus π‘š equals 60 and add 𝑛 minus π‘š equals 34, we get 𝑛 plus 𝑛 equals two 𝑛, 𝑛 minus π‘š equals zero, and 60 plus 34 is 94. Now we divide both sides by two to find that 𝑛 is 47. We now actually have three ways to determine π‘š. We can either plug 47 in to 𝑛 plus π‘š equals 60 or into 𝑛 minus π‘š equals 34. Or we can subtract these two equations to get an expression for two π‘š independent of the value of 𝑛. Just for demonstration, we’ll use the last method and subtract 𝑛 minus π‘š equals 34 from 𝑛 plus π‘š equals 60.

π‘š minus 𝑛 is zero. π‘š minus negative π‘š is π‘š plus π‘š, which is two π‘š. And 60 minus 34 is 26. So dividing both sides by two, we find that π‘š is equal to 13. Note that we found this value for π‘š without using the fact that we already knew that 𝑛 was 47, which means that we could’ve used this first to find that π‘š was 13 and then solved for 𝑛 afterward. Regardless of the order we choose when performing the calculations, we’ll find that 𝑛 is 47 and π‘š is 13.

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