# Question Video: Finding an Unknown by Evaluating Permutations Mathematics

Find the solution set of 42 (𝑥 + 3) 𝑃₃ = (𝑥 + 5) 𝑃₅.

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### Video Transcript

Find the solution set of 42 times 𝑥 plus three 𝑃 three equals 𝑥 plus five 𝑃 five.

The notation 𝑛𝑃𝑟 means the number of permutations of 𝑟 unique objects taken from a collection of 𝑛 unique objects. And it is calculated as 𝑛 factorial divided by 𝑛 minus 𝑟 factorial. The factorial of a positive integer 𝑛 is defined as the product of all of the positive integers from one to 𝑛 inclusive. And compactly, we can write this as 𝑛 factorial equals 𝑛 times 𝑛 minus one factorial. Using this recursive relationship for factorial, we can actually derive a similar recursive relationship for 𝑛𝑃𝑟.

This relationship is 𝑛𝑃𝑟 is equal to 𝑛 times 𝑛 minus one 𝑃𝑟 minus one. There are a number of ways to derive this identity. But the easiest way to see that it is true is to evaluate 𝑛 minus one 𝑃𝑟 minus one. This gives us 𝑛 minus one factorial divided by 𝑛 minus 𝑟 factorial. If we multiply by 𝑛, we have 𝑛 times 𝑛 minus one factorial divided by 𝑛 minus 𝑟 factorial. But our identity for factorials tells us that 𝑛 times 𝑛 minus one factorial is just 𝑛 factorial. So we have 𝑛 factorial divided by 𝑛 minus 𝑟 factorial, which is, of course, the definition of 𝑛𝑃𝑟. This identity will be quite useful in solving our equation.

We observe that 𝑥 plus five is two more than 𝑥 plus three and five is two more than three. So applying our identity once, we see that 𝑥 plus five 𝑃 five is equal to 𝑥 plus five times 𝑥 plus four 𝑃 four. If we apply the identity a second time, this time to 𝑥 plus four 𝑃 four, we see that 𝑥 plus five 𝑃 five is equal to 𝑥 plus five times 𝑥 plus four times 𝑥 plus three 𝑃 three, where we’ve substituted 𝑥 plus four times 𝑥 plus three 𝑃 three for 𝑥 plus four 𝑃 four using our identity. We can now equate this term which is equivalent to the right-hand side of our equation to the expression on the left-hand side.

Note the common factor of the nonzero integer 𝑥 plus three 𝑃 three on both sides of the equation. If we divide both sides by this number, we find that 42 equals 𝑥 plus five times 𝑥 plus four. We’ll solve this quadratic equation by relying on the fact that 𝑥 plus five and 𝑥 plus four are consecutive integers greater than one. Since 42 is the product of two consecutive integers greater than one, these integers are very close to the square root of 42. The square root of 42 is approximately 6.5, which is between six and seven. So our first guess is that 𝑥 plus four is six and 𝑥 plus five is seven, and indeed, seven times six is equal to 42.

Because these two products are equal, we conclude that 𝑥 plus four is six and 𝑥 plus five is seven, which means that 𝑥 is two. There is another solution for 𝑥, specifically when 𝑥 plus five is negative six and 𝑥 plus four is negative seven because negative six times negative seven is still positive 42. However, in this case, we would have 𝑥 equal to negative 11. But the notation 𝑛𝑃𝑟 comes with the implicit assumption that 𝑛 is greater than or equal to 𝑟. So it is a necessary assumption of our problem that 𝑥 plus five is greater than or equal to five because we have a permutation symbol with 𝑛 equal to 𝑥 plus five and 𝑟 equal five.

But this condition is equivalent to the condition that 𝑥 be greater than or equal to zero. So although this quadratic equation has two solutions, 𝑥 equals two and 𝑥 equals negative 11, the only solution that meets the additional constraint imposed by original equation that 𝑥 be greater than or equal to zero is the solution 𝑥 equals two. So although the solution set of the quadratic equation that we found has two elements, the solution set of our original equation has only a single element two.