Video: Using Properties of Combinations to Find an Unknown

At a college, there were 560 ways to select 13 students to attend a seminar. Determine the number of students at the college.

03:55

Video Transcript

At a college, there were 560 ways to select 13 students to attend a seminar. Determine the number of students at the college.

In this question, we’ve been given the total number of ways of choosing 13 students from the number that we have at the college. Let’s begin by defining the number of students we have at the college as 𝑛. And actually, what we’re working with here is combinations. This is a way of calculating the total outcomes of an event where the order of those outcomes doesn’t matter. And we say that the number of ways of choosing π‘Ÿ items from a total of 𝑛 items where the order doesn’t matter is 𝑛 choose π‘Ÿ, where 𝑛 choose π‘Ÿ is 𝑛 factorial over π‘Ÿ factorial times 𝑛 minus π‘Ÿ factorial.

Now, we’ve already defined 𝑛 as being the total number of students. We’re told that there are 560 ways to select 13 of these students. So, π‘Ÿ is 13, and then 𝑛 choose 13 is 560. Since 𝑛 choose 13 is 𝑛 factorial over 13 factorial times 𝑛 minus 13 factorial, we can form an equation in 𝑛. But how do we solve this equation? Well, we could imagine the expansion of 𝑛 factorial. We know it’s 𝑛 times 𝑛 minus one times 𝑛 minus two and so on. If we kept going all the way through to 𝑛 minus 13, we would be able to simplify by dividing through by 𝑛 minus 13 on the numerator and denominator of our fraction.

The problem is we then have 𝑛 times 𝑛 minus one times 𝑛 minus two all the way through to 𝑛 minus 12, still not a particularly nice equation to solve. But since our 𝑛 choose 13 isn’t a particularly big number, we’re simply going to use trial and error. We might begin by assuming that there are 14 students at the college and trying 𝑛 is equal to 14. The thing is, though, 𝑛 choose 𝑛 minus one is simply equal to 𝑛. So 14 choose 13 is equal to 14 not 560. So we know that 𝑛 is not equal to 14.

Instead, we’re going to try 𝑛 is equal to 15. Then, 15 choose 13 is 15 factorial over 13 factorial times 15 minus 13 factorial, which simplifies to 15 factorial over 13 factorial times two factorial. But of course, we can write 15 factorial as 15 times 14 times 13 and so on, or as 15 times 14 times 13 factorial. Then, we see we can divide both our numerator and denominator by 13 factorial. We also see that we can divide through by two, and this means that 15 choose 13 is simply 15 times seven divided by one. But that’s 105, which is still not equal to 560.

So next, we’ll try 𝑛 equals 16. So this time we’re going to work out 16 choose 13. It’s 16 factorial over 13 factorial times 16 minus 13 factorial. We write 16 minus 13 factorial as three factorial. And then we rewrite our numerator as the product of 16, 15, 14, and 13 factorial. Now, we can divide through by 13 factorial. And if we write three factorial as three times two times one, we see we can divide through by three and by two. So, 16 choose 13 becomes 16 times five times seven divided by one, which is 560. So, the number of ways of choosing 13 students from a total of 16 is 560 as required. This means 𝑛 is actually equal to 16. And there, therefore, must be 16 students at the college.

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