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Question Video: Using Properties of Combinations to Find an Unknown Mathematics • 12th Grade

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

04:47

Video Transcript

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

In this question, weโ€™ve been given the number of ways of choosing 13 students from the total number of students at a class. Letโ€™s begin by defining the total number of students we have at the class as ๐‘›. What we are dealing with in this question is combinations. This is a way of calculating the total outcomes of an event where the order of those outcomes doesnโ€™t matter. We say that the number of ways of choosing ๐‘Ÿ items from a total of ๐‘› items where the order doesnโ€™t matter is ๐‘› choose ๐‘Ÿ, where ๐‘› choose ๐‘Ÿ is equal to ๐‘› factorial divided by ๐‘Ÿ factorial multiplied by ๐‘› minus ๐‘Ÿ factorial.

In this question, weโ€™ve already defined ๐‘› as being the total number of students. We are told that there are 560 ways to select 13 of these students. So ๐‘Ÿ is equal to 13 and ๐‘› choose 13 is equal to 560. This means that ๐‘› factorial divided by 13 factorial multiplied by ๐‘› minus 13 factorial is equal to 560.

We now have an equation in ๐‘›. But how do we solve this? Well, we could imagine the expansion of ๐‘› factorial. We know itโ€™s ๐‘› multiplied by ๐‘› minus one multiplied by ๐‘› minus two and so on. If we kept going to ๐‘› minus 13, we would be able to simplify by dividing through by ๐‘› minus 13 factorial on the numerator and denominator of our fraction. The problem is we then have ๐‘› multiplied by ๐‘› minus one multiplied by ๐‘› minus two all the way through to ๐‘› minus 12, which is still not a particularly nice equation to solve. As a result, since our ๐‘› choose 13 value isnโ€™t a particularly big number, weโ€™re simply going to use trial and error or trial and improvement.

We might begin by assuming that there are 14 students at the class and trying ๐‘› is equal to 14. However, itโ€™s worth recalling that ๐‘› choose ๐‘› minus one is simply equal to ๐‘›. This means that 14 choose 13 is equal to 14 and not 560. Letโ€™s instead consider ๐‘› is equal to 15. Using the general formula 15 choose 13 is equal to 15 factorial divided by 13 factorial multiplied by two factorial. We know that we can write 15 factorial as 15 multiplied by 14 multiplied by 13 factorial. Dividing both the numerator and denominator by 13 factorial, we have 15 multiplied by 14 divided by two factorial. And since two factorial is equal to two, we have 15 multiplied by 14 divided by two, which is equal to 105. 15 choose 13 is 105 and not 560. If there were 15 students in the class, there would be 105 ways to select 13 students.

Next, we try ๐‘› is equal to 16. 16 choose 13 is equal to 16 factorial divided by 13 factorial multiplied by three factorial. The numerator can be simplified in a similar way. We can once again divide the top and bottom of our fraction by 13 factorial, leaving us with 16 multiplied by 15 multiplied by 14 divided by six. This is equal to 560.

We can therefore conclude that the number of ways of selecting 13 students from a total of 16 students is 560. And there are therefore 16 students at the class.

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