Lesson: When to use Combinations

Comment on When to use Combinations

ausm! thnx!! :)

Amazing explanation. Crisp and precise. I am glad that i found you guys for my prep.
greenlight-admin's picture

Glad you like it!

In the second question listed (urch webpage) I couldn't see the question even after scrolling to bottom. Could you look into that?
greenlight-admin's picture

Thanks for the heads up.
That site has had several problems recently, so I have deleted the link.

Cheers and thanks,

Combinations have always been terrible for me to understand. Thank you for the great explanation.

So how do we determine there are no different toppings, when the question states 3 different toppings which could be mushrooms, onions, pepperoni etc.
greenlight-admin's picture

You're referring to the question that starts 3:55 in the above video.

I'm not sure what you're asking when you say "So how do we determine there are no different toppings"

Can you please elaborate?


Superb explanation.

the problem -

How many 3 -digit number greater than 399 ending eith 5 or 7?

Kindly let me know if I can proceed as below steps:-
Let us take there are no restriction of 5 or 7 so
First digit - 6 ways
Second digit - 10 ways
Third digit - 10 ways.

So total no. of ways = 6 * 10 * 10 =600

Now let us assume the last digit is not filled by 5 or 7

First digit - 6 ways
Second digit - 10 ways
Third digit - 8 ways.

So total no. of ways = 6 * 10 * 8 =480 ways.

Therefore no. ways the number greater than 399 and the last digit is 5 or 7 = 600 -480 = 120 ways
greenlight-admin's picture

You're referring to the question that starts at 5:10 in the above video.

Yes, that approach works perfectly. It requires some extra steps, but the important thing is that you got the correct answer. Nice work!


Thanks Brent

Hello Brent,

Can you please clarify me, what difference in answer is actually observed when solving same question by fundamental counting principle as well as combination technique?
greenlight-admin's picture

If you're referring to the question that starts at 5:10 in the above video, it can't be solved using the combinations.

Here's my solution:

The question: How many 3-digit integers numbers greater than 399 end with 5 or 7?

Stage 1: Select the last (units) digit.
This digit must be 5 or 7
So, the stage can be completed in 2 ways

Stage 2: Select the first (hundreds) digit.
This digit must be 4, 5, 6, 7, 8 or 9
So, the stage can be completed in 6 ways

Stage 3: Select the second (tens) digit.
This digit must be 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9
So, the stage can be completed in 10 ways

By the fundamental counting principle, the total number of 3-digit number that satisfy the given conditions = (2)(6)(10) = 120


I can understand the logic behind the pizza question, with 12 choose 3. However, is there a faster way to solve this rather than multiplying 1 through 12, / 1 through 3 and 1 through 9 to give us 479,001,600/ 21,772,80= 220. It's quite time-consuming.
greenlight-admin's picture

The fast way looks like this:

12C3 = (12)(11)(10)/(3)(2)(1) = 220

Two videos from this video, (at https://www.greenlighttestprep.com/module/gre-counting/video/789) you'll learn how I calculated 12C3 above.


You know what's interesting about these combination questions? If you don't know whether to use the FCP or combination rule, try both and see what you get. For example, in the pizza example question, using the FCP gives you 1,320 while the combination rule gives you 220. This is very helpful in multiple choice questions because 1,320 is greatly spread out from 220. In almost all cases, such a widely spread out number from the rest of the numbers will be wrong. So if you don't know which method to use, and it's a multiple choice question, try both and see which answer matches. Now if this was a "provide the answer in the box" type of question, you're pretty much dead if you don't know whether to use combination or FCP. At that point, you need to guess one of the two answers you got.
greenlight-admin's picture

Good idea!

For the handshake example, I used a different strategy where I listed out the numbers 1 through 8, signifying each person. Then I drew these lines going from 1 to 2, 1 to 3, 1 to 4, etc. Repeat for the others without shaking hands with the same person (i.e. 2 to 3, 2 to 4, etc.). Then count up those handshakes. For person 1, there were 7 shakes, person 2, 6 shakes, etc. 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28. Same answer as given with the combination formula.
greenlight-admin's picture

That's the beauty of listing and counting!
If you list possible outcomes in a systematic way, you'll often see a pattern (e.g., 7 + 6 + 5 ....)

Hello - in 6:22. Why did you use 10 and not 9 ? I am trying to understand this. Thank you.
greenlight-admin's picture

The middle (tens) digit can be any digit (0,1,2,3,4,5,6,7,8 or 9)
So, there are 10 ways to complete that stage.

Hi Brent, it seems below page is no longer there? Is it still relevant? Thanks Brent

Page not found
The requested page "/articles/combinations-and-non-combinations-%e2%80%93-part-iibinations-%e2%80%93-part-ii" could not be found.
greenlight-admin's picture

Fixed! Thanks for the heads up.

Great thanks Brent.

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