GMAT Test Prep requires you know a thing or two about counting combinations and how they’re used on the GMAT. Watch the video below to learn more.
GMAT Test Prep Counting Combinations [Transcrip]
In this lesson, we’ll look at a quick way to calculate combinations. Now, here’s the combination formula. We can use it to determine the number of ways to select R objects from N unique objects when the order of the selected objects does not matter. Now while it’s perfectly fine to memorize this formula, there is a faster way to calculate combinations. To see how the faster technique works, let’s first calculate 8 choose 3. Here, N equals 8 and R equals 3. So when we plug these values into the formula, we get the following.
Our next step is to simplify 8 – 3 which equals 5. To evaluate this, we’ll expand the factorials and at this point we can cancel some of the numbers in the numerator and denominator. We’ll begin by cancelling the 5s, 4s, 3s, 2s, and 1s from the top and bottom. When we do this, we get the following. Now here comes a very important observation.
Notice that the numerator now consists of the first 3 terms of 8 factorial and the denominator is 3 factorial. It seems that the number 3 plays an important role in this calculation. But where does it come from? Well, it comes from up here. It is the value of R. We are selecting a set of 3 objects from 8 objects.
So to generalize the role that R plays in calculating combinations we can say that the value of N choose R will always be equal to the first R value of N factorial over R factorial. Now let’s continue with our calculations. We can further simplify this expression by recognizing that we can cancel the 6 in the numerator with the product 3 times 2 times 1 in the denominator. When we do this, we get 8 times 7 which equals 56. So 8 choose 3 is equal to 56.
Now let’s practice this technique. We’ll begin with 10 choose 2. Here, the R value is 2 so the numerator will be the first 2 values of 10 factorial and the denominator will be 2 factorial. Next, we’ll expand 2 factorial and we can simplify this to be 45. So we can select 2 objects from a group of 10 objects in 45 different ways.
Let’s try one more. How about 7 choose 3? Here the R value is 3 so the numerator will be the first 3 values of 7 factorial, and the denominator will be 3 factorial. Next we’ll expand 3 factorial. When we do this, we can see that we can cancel the 6 in the numerator with this product in the denominator to get 35. So, 3 objects can be selected from a group of 7 objects in 35 different ways.
Now before we end this lesson, I want to examine a special case. What happens if we have a combination question where we have N objects and we want to choose a 0 of them? For example, let’s say we have 8 employees and we must choose a 0 of them to be on a committee. In how many ways can we do this? Well, to find out we must evaluate 8 choose 0.
Now if we use our shortcut here, we run into a problem. When we apply this rule, we see that the numerator will be the product of the first 0 values of 8 factorial. So, what does this equal? Now the denominator equals 0 factorial and we learned in an earlier lesson that 0 factorial is equal to 1. So the denominator isn’t really a problem here. We do however have a problem with the numerator.
So how do we evaluate this? Well, perhaps we should apply some logic here. We have 8 employees and we must choose 0 of them. We can accomplish this task in one way. Nobody is on the committee. Now if that explanation feels inadequate, let’s evaluate 8 choose 0 using the original formula.
So, we’ll replace N and R with 8 and 0 to get the following. And now we can simplify this to get 8 factorial over 0 factorial times 8 factorial and since 0 factorial is equal to 1 we can see that 8 choose 0 evaluates to be 1. So in general we can say the N choose 0 will equal 1.
Okay, that concludes this lesson on a quick way to calculate combinations. Be sure to practice this technique so you can save time on test day.