Video: Evaluating Combinations to Find the Value of an Unknown Then Evaluating This Value in a Combination

Given that 𝑛Cβ‚„ = 5/2 𝑛, determine 𝑛C₃.

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Video Transcript

Given that 𝑛 choose four is equal to five over two 𝑛, determine 𝑛 choose three.

Let’s begin by recalling what we actually mean by this notation. 𝑛 choose π‘Ÿ is the number of ways of choosing our unique items from a total of 𝑛 in the collection, assuming that order doesn’t matter. It’s a combination. It’s 𝑛 factorial over π‘Ÿ factorial times 𝑛 minus π‘Ÿ factorial. Initially, we’re told that 𝑛 choose four is equal to five over two 𝑛. So let’s find an expression for 𝑛 choose four. By letting π‘Ÿ be equal to four, we find it’s equal to 𝑛 factorial over four factorial times 𝑛 minus four factorial. But of course this is equal to five over two 𝑛. So let’s form an equation in 𝑛. We get five over two 𝑛 equals 𝑛 factorial over four factorial times 𝑛 minus four factorial.

We can simplify this a little bit. We know that 𝑛 factorial is 𝑛 times 𝑛 minus one times 𝑛 minus two and so on. So we can rewrite it as 𝑛 times 𝑛 minus one times 𝑛 minus two times 𝑛 minus three times 𝑛 minus four factorial. And then we see that we’re going to be able to cancel a factor of 𝑛 minus four factorial from our numerator and denominator, leaving us with the equation five over two 𝑛 equals 𝑛 times 𝑛 minus one times 𝑛 minus two times 𝑛 minus three over four factorial. Let’s multiply both sides of our equation by four factorial, where four factorial is four times three times two times one, which is 24.

That means we get five over two times 24 which is 60𝑛 on the left-hand side. And on the right, we get 𝑛 times 𝑛 minus one times 𝑛 minus two times 𝑛 minus three. We subtract 60𝑛 from both sides, and we’re going to distribute our parentheses. By distributing 𝑛 times 𝑛 minus one times 𝑛 minus two times 𝑛 minus three, we get 𝑛 to the fourth power minus six 𝑛 cubed plus 11𝑛 squared minus six 𝑛. Let’s simplify by subtracting that 60𝑛 and the right-hand side of our equation now becomes 𝑛 to the fourth power minus six 𝑛 cubed plus 11𝑛 squared minus 66𝑛.

Now, to solve this equation for 𝑛, we’re going to need to factor. So we spot that we can begin by factoring 𝑛 from the right-hand side to get 𝑛 times 𝑛 cubed minus six 𝑛 squared plus 11𝑛 minus 66. But how do we factor 𝑛 cubed minus six 𝑛 squared plus 11𝑛 minus 66? Well, we begin by recalling the factor theorem. This says that if 𝑛 minus π‘Ž is a factor of the function 𝑓 of 𝑛, then 𝑓 of π‘Ž must be equal to zero. We can take a little guess as to what π‘Ž might be by looking for factors of negative 66. Let’s try π‘Ž is equal to six. 𝑓 of six is six cubed minus six times six squared minus 11 times six minus 66 which is equal to zero.

So 𝑓 of six is equal to zero, meaning 𝑛 minus six must be a factor of our cubic. This means we can perform polynomial long division to fully factor the cubic. 𝑛 cubed divided by 𝑛 is 𝑛 squared. Then, if we multiply 𝑛 squared by both parts of our binomial, we get 𝑛 cubed minus six 𝑛 squared. Subtracting these terms and we get zero. And now, we bring down the next two terms. That’s 11𝑛 minus 66. We’re going to divide 11𝑛 by 𝑛 to give us 11. And now, we multiply 11 by both terms in our binomial.

That gives us 11𝑛 minus 66. And subtracting, we see we get a remainder of zero, which we expected since 𝑛 minus six is a factor of our cubic. And so when we factor our equation, we get zero equals 𝑛 time 𝑛 minus six times 𝑛 squared plus 11. One of the solutions to this equation is 𝑛 equals zero. Another solution is when 𝑛 minus six is equal to zero, so 𝑛 is six. The equation 𝑛 squared plus 11 equals zero, however, has no real solutions because if we’re trying to solve it, we’d be finding the square root of negative 11. We know that in 𝑛 choose π‘Ÿ, 𝑛 must be a positive integer. So we choose the value 𝑛 equal six and disregard any further solutions.

Let’s clear some space and determine 𝑛 choose three if 𝑛 is equal to six. We’re going to evaluate six choose three. And so letting 𝑛 be equal to six and π‘Ÿ be equal to three in our earlier formula, we get six factorial over three factorial times three factorial. We write six factorial as six times five times four times three factorial. And then we cancel three factorial. But three factorial is six, so we can cancel another three factorial. And we find six choose three is simply equal to five times four which is 20. And so given that 𝑛 choose four is five over two 𝑛, we find 𝑛 choose three is equal to 20.

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