How to Do Combinations: A GRE and GMAT Introduction by Top-Score Tutor

Understanding combinations can seem overwhelming for many students. But, luckily, I am here to show you that they really aren’t that bad, and before you know it you will be finding combinations like it’s no-one’s business!

How might you be asked to count combinations? Well, a typical situation is this:
There are 8 employees in a team, and 3 of the employees are going to be promoted. In how many different ways can this be done?

Without some combinations knowledge, we are a little bit, how-to-say, royally screwed [sound effect]. We would have to list out the people, say A to H, and then find different groups of three until we think we have exhausted all the ways. Way too long! So, instead, mathematicians came up with a great time-saving formula:

n!/(n-r)!r!

I could give a bit more background to the origin of this formula if people ask for it in the comments below, but suffice to say if you know how to use it you will be greatly advantaged.

So, what does the n and r stand for, and what’s with the exclamation marks?
The n stands for the total from which we are picking out a group. It is always the biggest number. If you are selecting a group of 3 people from 8 people, 8 would be the n.

The r stands for the size of the group that you are picking out/selecting from the total. If you are selecting a team of 4 from a list of 9, the 4 would be your r.

The exclamation marks (‘!’) mean factorial. You multiply all the integers from 1 until you get to the number next to the exclamation mark. So, 5! = 5x4x3x2x1.

I will be doing another video on factorials specifically, but for the purposes of this formula, we just need to know that’s how they work.

I know you might think that the n – r and the r bit at the bottom is hard to remember, but an easy way to remember it is that the two terms on the bottom must add up to the term on the top. For example, with selecting 4 people from 9, we would have 4 and 5 factorial on the bottom and 9 factorial on top.
And when do we use this amazing formula? Whenever you need to know how many arrangements or combinations there are of doing something, when writing them all out would take too long.

You may have heard of another formula for ‘permutations’, which I will cover in another video, and wonder when to use the above combinations formula. The formula above is relevant for over 90% of the arrangement questions on the GRE and GMAT – it is by far the most important formula. To cut a long story short, we use it whenever the order of the people being selected does not matter.

Going back to our question, it does not matter if we promote persons BCD or CBD, we end up promoting the same three people.

Enough build-up, let’s use our incredible formula to answer our question!

The question shown is a classic combinations question. We don’t care about order, we just want to find all the ways of picking 3 people from a group of 8. If you want you can pause the video to see if you can use the formula yourself:

Here, n is the total, which is 8, and r is 3:
n!/(n-r)!r!= 8!/(8-3)!3!= 8!/5!3!=
8x7x6x5x4x3x2x1/((5x4x3x2x1)(3x2x1))= 8x7x6x5x4x3x2x1/((5x4x3x2x1)(3x2x1))
= 8x7x6/((3x2x1) ) = 8x7/1 = 56.

Notice how much always cancels out with factorials!

Isn’t this so exciting? We have saved so much time and got the right answer. If this is starting to make sense do hit the like button and subscribe! Also, do comment down below with any gratitude or any questions you may have.

Before you go, let me give you one more example of this core technique, just to bed it in.

Imagine there are 6 women and 5 men just chilling out relaxing, having some fun. We want to select 2 women and 2 men for a neighbourhood event.

How many ways are there of selecting 2 women and 2 men from the 11 people available?
Pause the video and have a go if you wish. Otherwise, listen-up.

First, we pick the women. Do not lump everyone together, we have to do things step-by-step!
There are 6 women available and we are picking 2. Our n is 6 and our r is 2:

n!/(n-r)!r!= 6!/(6-2)!2!= 6!/4!2!
= 6x5x4x3x2x1/((4x3x2x1)(2x1))= 6x5/2x1=15
n!/(n-r)!r!= 5!/(5-2)!2!= 5!/3!2!
= 5x4x3x2x1/((3x2x1)(2x1))= 5x4/2x1=10

Finally, what do we do with these two results – 15 selections possible of women, 10 selections possible with men?

Well, given that we want 2 women AND 2 men, we multiply the 15 and 10 to get 150 total possibilities. AND always means multiply in probability and combinations.

There are 150 (15 x 10) ways we can select a team of 2 women and 2 men from 6 women and 5 men.

Now, these two examples encapsulate the most common ways you will be tested on combinations in the GRE and GMAT. I will not pretend that it covers every single question type, that would be a long video. But, if this video gets a good enough response, I will definitely work my way through the other ways too!

Thank you so much for watching and have a great day!

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