Post your question in the Comment section below, and a GRE expert will answer it as fast as humanly possible.

- Video Course
- Video Course Overview
- General GRE Info and Strategies - 7 videos (free)
- Quantitative Comparison - 7 videos (free)
- Arithmetic - 42 videos
- Powers and Roots - 43 videos
- Algebra and Equation Solving - 78 videos
- Word Problems - 54 videos
- Geometry - 48 videos
- Integer Properties - 34 videos
- Statistics - 28 videos
- Counting - 27 videos
- Probability - 25 videos
- Data Interpretation - 24 videos
- Analytical Writing - 9 videos (free)
- Sentence Equivalence - 39 videos (free)
- Text Completion - 51 videos
- Reading Comprehension - 16 videos

- Study Guide
- About
- Office Hours
- Extras
- Prices

## Comment on

When to use Combinations## ausm! thnx!! :)

## Amazing explanation. Crisp

## Glad you like it!

Glad you like it!

## Hi!

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

Thanks

## Thanks for the heads up.

Thanks for the heads up.

That site has had several problems recently, so I have deleted the link.

Cheers and thanks,

Brent

## Combinations have always been

## So how do we determine there

## You're referring to the

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?

Cheers,

Brent

## 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

so,

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

## You're referring to the

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!

Cheers,

Brent

## 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?

## If you're referring to the

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

Cheers,

Brent

## I can understand the logic

## The fast way looks like this:

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.

Cheers,

Brent

## You know what's interesting

## Good idea!

Good idea!

## For the handshake example, I

## That's the beauty of listing

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

## The middle (tens) digit can

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.