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

Given that 𝑛C₃ = 2𝑛, evaluate 𝑛C₃.

03:25

Video Transcript

Given that 𝑛 choose three is equal to two 𝑛, evaluate 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 π‘Ÿ items from a total collection of 𝑛 items when order doesn’t matter. It’s a combination. 𝑛 choose π‘Ÿ or 𝑛Cπ‘Ÿ is evaluated by dividing 𝑛 factorial by π‘Ÿ factorial times 𝑛 minus π‘Ÿ factorial. This means we can form an expression for 𝑛 choose three. By letting π‘Ÿ be equal to three, we get 𝑛 factorial over three factorial times 𝑛 minus three factorial.

We’re told this is equal to two 𝑛. So let’s form an equation in 𝑛. We get 𝑛 factorial over three factorial times 𝑛 minus three factorial. But, of course, by the definition of the factorial, we know 𝑛 factorial is 𝑛 times 𝑛 minus one times 𝑛 minus two, and so on. This means we can write it as 𝑛 times 𝑛 minus one times 𝑛 minus two times 𝑛 minus three factorial.

And so, if we rewrite our numerator in this form, we see that we can divide through by a constant factor of 𝑛 minus three factorial such that 𝑛 times 𝑛 minus one times 𝑛 minus two over three factorial is equal to two 𝑛.

Let’s solve this equation for 𝑛 by first multiplying through by three factorial. Now, three factorial is three times two times one. So it’s actually six. And we get 𝑛 times 𝑛 minus one times 𝑛 minus two equals 12𝑛. Next, we’re going to distribute our parentheses. We get 𝑛 times 𝑛 squared minus three 𝑛 plus two equals 12𝑛. Then we distribute further to get 𝑛 cubed minus three 𝑛 squared plus two 𝑛 equals 12𝑛.

Now, we have a cubic. So we’re going to subtract 12𝑛 from both sides to get this equal to zero. And we get 𝑛 cubed minus three 𝑛 squared minus 10𝑛 equals zero. Which we can solve for 𝑛 by factoring first by removing the 𝑛 and then factoring the quadratic 𝑛 squared minus three 𝑛 minus 10 to get 𝑛 plus two times 𝑛 minus five.

The roots of this equation are the values of 𝑛 such that 𝑛 is equal to zero, 𝑛 plus two is equal to zero, or 𝑛 minus five is equal to zero. That’s 𝑛 equals zero, 𝑛 equals negative two, and 𝑛 equals five. Now, in fact, 𝑛 must be a positive integer. So we disregard two of the solutions, and we get 𝑛 is equal to five.

And we’re now able to evaluate 𝑛 choose three. In this case, it’s actually five choose three. Going back to our definition for 𝑛 choose π‘Ÿ, we’re going to let 𝑛 be equal to five and π‘Ÿ be equal to three. That gives us five factorial over three factorial times five minus three factorial, which is five factorial over three factorial times two factorial.

This time, we’re going to write our numerator as five times four times three factorial. Allowing us to divide through by a constant factor of three factorial. But also two factorial is just two. So we’re going to divide through by two. And we see that five choose three is five times two, which is simply 10.

And so given that 𝑛 choose three is equal to two 𝑛, we find 𝑛 choose three is equal to 10.

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