Question Video: Evaluating Permutations to Find the Value of an Unknown Mathematics

Find the value of π‘₯ given that 235𝑃π‘₯ βˆ’ 3π‘₯ 235𝑃(π‘₯ βˆ’ 1) = 0.

03:30

Video Transcript

Find the value of π‘₯ given that 235𝑃π‘₯ minus three π‘₯ times 235𝑃π‘₯ minus one equals zero.

In order to find π‘₯, we need to understand the notation used in this equation. The notation π‘›π‘ƒπ‘Ÿ means the number of permutations of our unique items taken from a collection of 𝑛 unique items. And we can calculate this as 𝑛 factorial divided by 𝑛 minus π‘Ÿ factorial. The factorial of a positive integer 𝑛 is the product of all the integers from one to 𝑛 inclusive.

It follows directly from this definition that 𝑛 factorial is equal to 𝑛 times 𝑛 minus one factorial. Equivalently, 𝑛 factorial is 𝑛 times 𝑛 minus one times 𝑛 minus two factorial and so on. Expanding out factorials like this is very helpful in simplifying expressions.

In our equation, the two elements we need to simplify are 235𝑃π‘₯ and 235𝑃π‘₯ minus one. Note how similar these are. The only difference is that their π‘Ÿ differs by one. When we see something like this, it’s usually the case that one term is a simple multiple of the other term. And this means that we can simplify our expression by figuring out what that multiple is. So, we’re looking in general to find a number such that π‘›π‘ƒπ‘Ÿ is that number times π‘›π‘ƒπ‘Ÿ minus one. We’ve written the relationship this way because π‘›π‘ƒπ‘Ÿ is always greater than or equal to π‘›π‘ƒπ‘Ÿ minus one.

Anyway, we can see that the number that we’re looking for is π‘›π‘ƒπ‘Ÿ divided by π‘›π‘ƒπ‘Ÿ minus one. Expanding using our definition of π‘›π‘ƒπ‘Ÿ in terms of factorials, we get 𝑛 factorial divided by 𝑛 minus π‘Ÿ factorial all divided by 𝑛 factorial divided by 𝑛 minus π‘Ÿ plus one factorial. 𝑛 factorial in the numerator of the numerator divided by 𝑛 factorial in the numerator of the denominator is one. And 𝑛 minus π‘Ÿ factorial in the denominator of the numerator goes into the denominator of the overall fraction. And 𝑛 minus π‘Ÿ plus one factorial in the denominator of the denominator goes into the numerator of the overall fraction.

This leaves us with 𝑛 minus π‘Ÿ plus one factorial divided by 𝑛 minus π‘Ÿ factorial. We can now simplify using our expression for 𝑛 factorial. This expression tells us that 𝑛 minus π‘Ÿ plus one factorial is equal to 𝑛 minus π‘Ÿ plus one times 𝑛 minus π‘Ÿ factorial. Noting the common factor of 𝑛 minus π‘Ÿ factorial in the numerator and denominator, we see that this expression is simply equal to 𝑛 minus π‘Ÿ plus one. So, the number that we’re looking for in our relationship is 𝑛 minus π‘Ÿ plus one. Now, we’ll use this general relationship to substitute in for 235𝑃π‘₯ in our original equation.

Replacing 𝑛 with 235 and π‘Ÿ with π‘₯, we have 235 minus π‘₯ plus one times 235𝑃π‘₯ minus one minus three π‘₯ times 235𝑃π‘₯ minus one equals zero. Now, both terms on the left-hand side have a common factor. Since this common factor is not zero, we can divide both sides of the equation by 235𝑃π‘₯ minus one to get rid of it. Our equation then simplifies to 235 minus π‘₯ plus one minus three π‘₯ equals zero. 235 plus one is 236, and negative π‘₯ minus three π‘₯ is negative four π‘₯. Adding four π‘₯ to both sides, we find that 236 is equal to four π‘₯. And finally dividing both sides by four, we arrive at our answer π‘₯ equals 59.

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