Question Video: Evaluating Combinations to Find the Value of an Unknown

If 𝑛C₃ = 120, find 𝑛.

05:22

Video Transcript

If 𝑛 choose three is equal to 120, find 𝑛.

We begin by recalling the definition of 𝑛Cπ‘Ÿ. It tells us the number of ways we can choose π‘Ÿ unique items from a collection of 𝑛 items, where order doesn’t matter. And it’s defined as 𝑛 factorial over π‘Ÿ factorial times 𝑛 minus π‘Ÿ factorial. 𝑛Cπ‘Ÿ or 𝑛 choose π‘Ÿ is sometimes alternatively written as shown. Now, in our question, we’re told the value of 𝑛 choose three. So let’s rewrite 𝑛 choose three using our notation.

We replace π‘Ÿ with three and we get 𝑛 factorial over three factorial times 𝑛 minus three factorial. Of course, we’re told that this is equal to 120. So, we can now say that 𝑛 factorial over three factorial times 𝑛 minus three factorial equals 120. Let’s simplify it a little bit by multiplying both sides by three factorial. And so, we get 𝑛 factorial over 𝑛 minus three factorial equals 120 times three factorial. But, of course, three factorial is three times two times one, which is six. And since 120 times six is 720, we get 𝑛 factorial over 𝑛 minus three factorial equals 720.

Now, we apply the definition of factorial. We know 𝑛 factorial is 𝑛 times 𝑛 minus one times 𝑛 minus two, and so on. Since this continues, we can write it as 𝑛 times 𝑛 minus one times 𝑛 minus two times 𝑛 minus three factorial. And then, we see that since we’re dividing 𝑛 factorial by 𝑛 minus three factorial, these factorials will cancel out. And our equation simplifies really nicely. We have 𝑛 times 𝑛 minus one times 𝑛 minus two equals 720. Essentially, we have a cubic. So, we’re going to distribute the parentheses and set this equal to zero.

By multiplying 𝑛 minus one by 𝑛 minus two, we get 𝑛 squared minus three 𝑛 plus two. And then we multiply by 𝑛 and the left-hand side becomes 𝑛 cubed minus three 𝑛 squared plus two 𝑛 equals 720. Remember, to solve, we need to set this equal to zero and factor. We could use a polynomial solver on a calculator, but let’s recall the method that allows us to factor cubics. We can use 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 will be equal to zero. We know that our value of π‘Ž will be some factor of negative 720.

So, let’s try π‘Ž equals 10. That’s a fairly obvious factor of negative 720. If 𝑛 minus 10 is a factor, then 𝑓 of 10, which is 10 cubed, minus three times 10 squared plus two times 10 minus 720 will be equal to zero. Well, in fact it is. So, 𝑛 minus 10 is a factor of 𝑛 cubed minus three 𝑛 squared plus two 𝑛 minus 720. Let’s use polynomial long division to divide our function by 𝑛 minus 10. 𝑛 cubed divided by 𝑛 is 𝑛 squared. We then multiply 𝑛 squared by each term in our binomial. That gives us 𝑛 cubed minus 10𝑛 squared. Next, we subtract these terms. 𝑛 cubed minus 𝑛 cubed is zero. Then, negative three 𝑛 squared minus negative 10𝑛 squared is seven 𝑛 squared. We bring the remaining two terms down. And then, we divide seven 𝑛 squared by 𝑛. That gives us seven 𝑛.

Now, we multiply seven 𝑛 by the two terms in our binomial, giving us seven 𝑛 squared minus 70𝑛. Once again, we subtract. Seven 𝑛 squared minus seven 𝑛 squared is zero. Then, two 𝑛 minus negative 70𝑛 is 72𝑛. We bring the negative 720 down and then divide 72𝑛 by 𝑛, giving us 72. Once again, we multiply 72 by the two terms in our binomial, giving us 72𝑛 minus 720. Meaning that, as we expected, we have a remainder of zero. So, we get 𝑛 minus 10 times 𝑛 squared plus seven 𝑛 plus 72 equals zero.

Our job now is to check whether 𝑛 squared plus seven 𝑛 plus 72 is factorable. We’re going to use the discriminant. The discriminant of a quadratic in the form π‘Žπ‘› squared plus 𝑏𝑛 plus 𝑐 equals zero is 𝑏 squared minus four π‘Žπ‘. Here, π‘Ž, which is the coefficient of 𝑛 squared, is one; 𝑏, which is the coefficient of 𝑛, is seven; and 𝑐, which is the constant, is 72. The discriminant is, therefore, seven squared minus four times one times 72. Well, this is less than zero.

This tells us the equation 𝑛 squared plus seven 𝑛 plus 72 equals zero has no real solutions, and that means it’s not factorable. It also means the only solution to the cubic, 𝑛 cubed, minus three 𝑛 squared plus two 𝑛 minus 720 equals zero, at least the only real solution, is when 𝑛 minus 10 is equal to zero. By adding 10 to both sides, we find 𝑛 is equal to 10.

And so, if 𝑛 choose three is 120, 𝑛 must be equal to 10. Of course, we could check this by going back to our definition for 𝑛 choose π‘Ÿ and checking that 10 choose three does indeed give us 120.

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