Lesson Video: Properties of Permutations | Nagwa Lesson Video: Properties of Permutations | Nagwa

Lesson Video: Properties of Permutations Mathematics

In this video, we will learn how to use the properties of permutations to simplify expressions and solve equations.

16:33

Video Transcript

Properties of Permutations

In this video, we will learn how to use properties of permutations to simplify expressions and solve equations. First, let’s consider a few things we know about permutations. A permutation is an arrangement of a collection of items where order matters and repetition is not allowed. You might hear this said as permutations represent counting without replacement in which order matters. We calculate the number of possible permutations for a collection of items with this formula, 𝑛𝑃𝑟 equals 𝑛 factorial over 𝑛 minus 𝑟 factorial. Given nonnegative integers 𝑛 and 𝑟 satisfying that 𝑛 is greater than or equal to 𝑟, the permutation 𝑛𝑃𝑟 represents the number of different ways to order 𝑟 objects from a total of 𝑛 distinct objects.

It’s also important to note other common notations used for 𝑛𝑃𝑟. You might have superscript 𝑛 𝑃 subscript 𝑟, or 𝑃 superscript 𝑛 subscript 𝑟, or 𝑃 subscript 𝑛 comma 𝑟, or 𝑃 𝑛, 𝑟. Each of these notations represent the number of ways we can order 𝑟 elements from a set of 𝑛 elements with no repetition. Now, let’s consider some properties that all permutations share. For a permutation 𝑛𝑃𝑟, it is equal to 𝑛 times the permutation of 𝑛 minus one 𝑃 𝑟 minus one.

If we consider this property for the permutation five 𝑃 three, we should find that it is equal to five times the permutation four 𝑃 two. We know that 𝑛𝑃𝑟 equals 𝑛 factorial over 𝑛 minus 𝑟 factorial. And so we can rewrite five 𝑃 three as five factorial over two factorial. Additionally, we can rewrite four 𝑃 two as four factorial over two factorial. By expanding the factorials on either side of the equation, we find out that this is true. Both five 𝑃 three and five times four 𝑃 two are equal to 60.

A related property of factorials that we often use when dealing with permutations is 𝑛 factorial is equal to 𝑛 times 𝑛 minus one factorial, which means six factorial is equal to six times six minus one factorial. Six factorial is equal to six times five factorial. We can extend this property further to say 𝑛 factorial is equal to 𝑛 times 𝑛 minus one times 𝑛 minus two factorial. If we use our six factorial example again, we can show that it is equal to six times five times four factorial. Before we move on, we should also note that zero factorial equals one, and one factorial also equals one.

We can now consider some examples where we use the definition of permutation and its properties to solve equations.

Evaluate 123𝑃10 over 122𝑃 nine.

We’ve been given a fraction in which the numerator and the denominator are both permutations in the form 𝑛𝑃𝑟. One way to solve this problem would be to expand the numerator and the denominator. The numerator 123𝑃10 becomes 123 factorial over 123 minus 10 factorial. And then, we would divide that by 122 factorial divided by 122 minus nine factorial. Solving the subtraction, we can simplify both of these expressions to 113 factorial. We know that dividing by a fraction is the same thing as multiplying by the reciprocal. And our 113 factorial in the numerator and the denominator cancel each other out. To do a bit more simplifying, we can say that 𝑛 factorial is equal to 𝑛 minus one factorial. And that means we can rewrite 123 factorial as 123 times 122 factorial. In this form, we’re able to quickly see that the 122 factorial in the numerator and the denominator cancel out, leaving us with 123.

However, there was a more straightforward way to solve this problem using a property of permutations. This tells us for any permutation 𝑛𝑃𝑟, it will be equal to 𝑛 times 𝑛 minus one 𝑃 𝑟 minus one. Therefore, 123𝑃10 is equal to 123 times 122𝑃 nine. Using this method of simplification, we’d be able to remove the 122𝑃 nine from the numerator and the denominator without expanding both of them to factorial form. In either case, we see that 123𝑃10 over 123𝑃 nine is equal to 123.

Let’s look at another example.

If 𝑛𝑃15 equals 23 times 𝑛 minus one 𝑃14, find 𝑛.

In our equation, we have a factorial that is equal to 23 times another factorial. This permutation form tells us how many ways we can order 𝑟 elements from a set of 𝑛 elements with no repetition. On the left-hand side of the equation, we have 𝑛 elements. And on the right-hand side, we have 𝑛 minus one elements. The number of elements we’re selecting, 𝑟, on the left-hand side is 15. And that means we could say that on the right-hand side, we have 𝑟 minus one, which is 14. This means that our equation fits the form 𝑛𝑃𝑟 is equal to 𝑛 times 𝑛 minus one 𝑃 𝑟 minus one. And 𝑛 must be equal to 23. Plugging in what we know, we get 23𝑃15 is equal to 23 times 22𝑃14.

We can check that these values are equivalent by using the permutation function on a calculator or a computer. And then, we’ve shown that here 𝑛 is equal to 23. Moving on to our third example.

If 49𝑃 𝑟 plus three equals 34 times 49𝑃 𝑟 plus two, find the value of 𝑟 minus six factorial.

One way to solve this will be to first convert each of these permutations into factorial form using the definition 𝑛𝑃𝑟 equals 𝑛 factorial over 𝑛 minus 𝑟 factorial. We rewrite 49𝑃 𝑟 plus three as 49 factorial over 49 minus 𝑟 plus three factorial. Be careful to notice that in this case, our 𝑟, the number of elements we’re choosing, is 𝑟 plus three and that the subtraction applies to both the 𝑟 and the three. The other side of the equation becomes 34 times 49 factorial over 49 minus 𝑟 plus two factorial. We have 49 factorial in the numerator on both sides of the equation, which means we can multiply both sides of the equation by one over 49 factorial.

On the left, we’ll be left with the numerator of one. And in the denominator, we can subtract 𝑟 and subtract three from 49, which will give us one over 46 minus 𝑟 factorial. The numerator on the right side becomes 34, and the denominator is 47 minus 𝑟 factorial. At this point, it might not seem clear what we should do. But since we’re trying to solve for 𝑟, it’s a good idea to get both 𝑟-values on the same side of the equation. So we’ll multiply both sides of the equation by 47 minus 𝑟 factorial. This gives us 47 minus 𝑟 factorial over 46 minus 𝑟 factorial equals 34. Our denominator is a factorial that is one less than our numerator.

We know that 𝑛 factorial is equal to 𝑛 times 𝑛 minus one factorial. We can apply this here, but we need to be careful. In our case, the 𝑛 is 47 minus 𝑟. And that means 47 minus 𝑟 factorial is equal to 47 minus 𝑟 times 47 minus 𝑟 minus one factorial. We can simplify 47 minus 𝑟 factorial to be equal to 47 minus 𝑟 times 46 minus 𝑟 factorial. If we substitute this in for our numerator, the 46 minus 𝑟 factorial cancels out in the numerator and the denominator. We then have 47 minus 𝑟 equals 34. If we subtract 47 from both sides, we find that negative 𝑟 equals negative 13, which makes 𝑟 equal to positive 13. Our final step will be to calculate 𝑟 minus six factorial, which we now know is 13 minus six factorial. Solving this by hand or with a calculator will yield that seven factorial equals 5,040.

In our next example, we’ll look at an equation that has factorials and a permutation in the equation and solve for a missing value.

If 𝑥 minus 47 factorial times 𝑥𝑃47 equals 3,906 times 𝑥 minus two factorial, find the value of 𝑥.

If we look at the equation we’re given, we have an 𝑥-value and two different factorials, and we have the 𝑥 as the set from which we’re choosing a permutation. We know that for any factorial in the form 𝑛𝑃𝑟, we can rewrite it as 𝑛 factorial over 𝑛 minus 𝑟 factorial, which means we can rewrite 𝑥𝑃47 as 𝑥 factorial over 𝑥 minus 47 factorial. If we bring down the rest of the equation, we see that we have 𝑥 minus 47 factorial in the numerator and the denominator on the left-hand side. Therefore, our equation simplifies to 𝑥 factorial equals 3,906 times 𝑥 minus two factorial.

As we’re trying to solve for the value of 𝑥, we want to get both of our 𝑥-values on the same side of the equation. To do that, we divide the left- and right-hand side of the equation by 𝑥 minus two factorial. And we have 𝑥 factorial over 𝑥 minus two factorial equals 3,906. At this point, it seems like it’s not possible to simplify further. But remember that 𝑛 factorial can be rewritten as 𝑛 times 𝑛 minus one factorial. In fact, you can extend this property even further to something like 𝑛 factorial is equal to 𝑛 times 𝑛 minus one times 𝑛 minus two factorial.

Looking at our numerator, 𝑥 factorial, it can be rewritten and expanded as 𝑥 times 𝑥 minus one times 𝑥 minus two factorial. And then, the term 𝑥 minus two factorial occurs in both the numerator and the denominator and therefore can be canceled, leaving us with 𝑥 times 𝑥 minus one equals 3,906. In any permutation, the 𝑛-value must be a nonnegative integer. This means our 𝑥-value is an integer. We’re looking for some integer 𝑥 and then the integer that is one less than that, which multiply together to equal 3,906. One strategy for estimating this is to use your calculator and find the square root of 3,906, which is about 62.5, rounding to the nearest integer 62. We then divide 3,906 by 62 on the calculator, and we get 63. Since 63 times 62 equals 3,906, we can say that 𝑥 equals 63.

Another way to solve this problem would be to create a quadratic equation by multiplying 𝑥 times 𝑥 minus one. And then setting this equation equal to zero, you could solve using the quadratic formula or by factoring. If you solve the question using the quadratic formula, it’s important to note that you will find 𝑥 equals two different values, and one of those values will be a negative. Solving with the quadratic formula will yield 𝑥 equals 63 and 𝑥 equals negative 62. The negative 62 is not possible because the value of 𝑥 here must be a nonnegative integer. We cannot have a negative 𝑛-value because a permutation is the number of ways to choose 𝑟 objects from a set of 𝑛 distinct objects. This makes the only valid option for 𝑛, 63. You could also check this value is true by plugging 63 back into the original equation.

To finish, let’s review the key points. The number of permutations of size 𝑟 taken from a set of size 𝑛 is given by 𝑛𝑃𝑟 equals 𝑛 factorial over 𝑛 minus 𝑟 factorial. Using the property of factorials that 𝑛 factorial equals 𝑛 times 𝑛 minus one factorial, we can simplify fractional expressions involving permutations. And finally, using the property of permutations that 𝑛𝑃𝑟 equals 𝑛 times 𝑛 minus one 𝑃 𝑟 minus one, we can simplify some expressions involving permutations.

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