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## Comment on

Introduction to Combinations## Dear Brent

I solved the "Will must choose a 3-character computer password, consisting of 1 letter from the alphabet and 2 distinct digits" question in the following way:

1L 2D

The letter can be chosen out in 26 ways

The digits in 10 x 9 (since the two must be distinct)

And the letter can take one of 3 places within the code.

Therefore, it can be achieved in 26 x 10 x 9 x 3 ways.

(I know the math turns out to be the same, but is the approach wrong? )

## Question link: https:/

Question link: https://gmatclub.com/forum/will-must-choose-a-3-character-computer-passw...

That's a perfectly valid approach - great work!

## Can you pls help me

S is a set of points in the plane (IT DOES NOT MENTION IF THE POINTS ARE COLLINEAR OR DISPERSED). How many distinct triangles can be drawn that have three of the points in S as vertices?

(1) The number of distinct points in S is 5. - SO WHY IS (1) NOT SUFFICIENT ON ITS OWN?

(2) No three of the points in S are collinear.

## Question link: https:/

Question link: https://gmatclub.com/forum/s-is-a-set-of-points-in-the-plane-how-many-di...

If it isn't stated whether the points are collinear or dispersed, then we can't make any conclusion about their location.

So, when statement 1 tells us that there are 5 points, we have no idea whether the points are collinear or dispersed. Let's examine two possible cases.

Case a - All 5 points are collinear. In this case, we cannot create ANY triangles

Case b - The 5 points are dispersed (no 3 points are collinear). In this case, we can create 10 triangles, since any 3 selected points will create a triangle. 5C3 = 10

Does that help?

## How do we identify if the

## I'm not sure what you mean.

I'm not sure what you mean. Combination questions are a type of counting question. So, if a question is a combination question, then it is also a counting question.

If you're looking for guidance regarding when we need to use combinations and when we need to use a different approach, I cover that in the following video: https://www.gmatprepnow.com/module/gmat-counting/video/788

I've also written 3 articles on that topic:

Article #1: https://www.gmatprepnow.com/articles/combinations-and-non-combinations-%...

Article #2: https://www.gmatprepnow.com/articles/combinations-and-non-combinations-%...

Article #3: https://www.gmatprepnow.com/articles/does-order-matter-combinations-and-...

## Brent,

In the problem below, on the second criteria... what is the quickest way to come up with finding 6 choose 2 or 4 other than trying out each one. This can take more than 2 minutes in the exam, unless you know (as your explanation goes).

https://gmatclub.com/forum/from-a-group-of-6-employees-k-employees-are-chosen-to-be-on-the-party-237543.html

Thanks

Sri

## Question link: https:/

Question link: https://gmatclub.com/forum/from-a-group-of-6-employees-k-employees-are-c...

Great question, Sri.

An interesting property of combinations is that nCr = nC(n-r)

Some examples:

10C3 = 10C7

5C1 = 5C4

7C5 = 7C2

I'll use an illustrative example to show WHY this is the case.

Let's say you have 6 friends, and you can choose 4 of them to attend your party. We can do this in 6C4 ways

However, we can think of this situation in a different way. Rather than select 4 friends (out of 6 friends) to attend your party, we could just select 2 friends (out of 6 friends) to NOT attend the party, and then the remaining 4 friends get to attend the party. Notice that the outcome is exactly the same.

We can select 2 friends from 6 friends in 6C2 ways.

Since the outcome for each approach is the same, it must be the case that 6C4 = 6C2

Now to your question!

If I have the option of calculating 6C4 or 6C2, it's faster to calculate the one that has the smaller second number.

So, we can calculate 6C2 faster than we can calculate 6C4

6C2 = (6)(5)/(2)(1) = 15 (for more, see https://www.gmatprepnow.com/module/gmat-counting/video/789)

Does that help?

Cheers,

Brent

## How do we determine when to

Can you please break this down it is really confusing to me deciding what question I should use FCP for and which one to use the combination formula for.

Thank you

## We have a video that

We have a video that specifically answers your question: https://www.gmatprepnow.com/module/gmat-counting/video/788

Cheers,

Brent

## Hi Brent, facing some

https://gmatclub.com/forum/will-must-choose-a-3-character-computer-password-consisting-of-1-lett-223186.html

## Question link: https:/

Question link: https://gmatclub.com/forum/will-must-choose-a-3-character-computer-passw...

In Stages 1 & 2, we select the 3 letters that we will be using.

For example, during those two stages, we might select two W's and one Q

At this point, we have NOT YET arranged the three letters.

In Stage 3, we arrange the three selected characters.

Cheers,

Brent

## Hi Brent,

Have a question, these two questions below look exactly similar to me and I followed the same approach (FCP) to answer both. While I got the answer correct for one (triangle) and not for the other (Rectangle). Could you help me understand why FCP approach doesn't apply to Rectangle one?

https://gmatclub.com/forum/right-triangle-pqr-is-to-be-constructed-in-the-xy-plane-so-71597.html

https://gmatclub.com/forum/rectangle-abcd-is-constructed-in-the-coordinate-plane-parall-142272.html

## Question links:

Question links:

1) https://gmatclub.com/forum/right-triangle-pqr-is-to-be-constructed-in-th...

2) https://gmatclub.com/forum/rectangle-abcd-is-constructed-in-the-coordina...

I have a feeling that, for question #2, you counted some rectangles more than once.

If you can show me your solution to the rectangle question, I should be able to pinpoint the problem.

Cheers,

Brent

## Sure! In the rectangle ABCD

A - x1 y1

B - x2 y1

C - x2 y2

D - x1 y2

So we need to find two values of x and two values of y

x1 and x2 can have 9*8 = 72 ways

y1 and y2 can have 11*10 = 110 ways, totaling into 7920 ways.

## This solution has the same

This solution has the same issue as your solution below (regarding the 4 pairs of twins).

You wrote: x1 and x2 can have 9*8 = 72 ways

One possible outcome is that we choose 4 for x1 and choose 10 for x2

Another possible outcome is that we choose 10 for x1 and choose 4 for x2

These two outcomes are IDENTICAL.

Since the order in which we select values for x1 and x2 does not matter, we should use combinations.

We can select 2 x-values from 9 possible values in 9C2 ways (36 ways)

The same applies to selecting the y-values.

Does that help?

Cheers,

Brent

## Understood. So why does the

## Notice that a rectangle in

Notice that a rectangle in this question can be defined by any two points on the diagonal of the rectangle.

For example, if I tell you that (4, 7) and (2, 5) are points on the diagonal of a rectangle, then we automatically know the other two points will be (4, 5) and (2, 7).

The same cannot be said of the right triangles in the other question.

Cheers,

Brent

## Hi Brent,

In the question below, I followed a different approach. Here it is - 4 sets of twin = 8 people overall

Stage 1: select the 1st member in 8C1 = 8 ways

Stage 2: Supposing that the selected member is from Set 1 of the 4 sets of twins, total number of people to choose from becomes 6, therefore select the 2nd member in 6C1 = 6 ways

Stage 3: similarly, select the 3rd member in 4C1 = 4 ways.

Wanted to understand what's wrong with this approach since the restricted is followed.

https://www.beatthegmat.com/combinations-t123249.html

## Question link: https://www

Question link: https://www.beatthegmat.com/combinations-t123249.html

In your solution, you have counted each outcomes 6 times.

Let's say that a pair twins is represented by uppercase and lowercase versions of the same letter (e.g., B and b)

So, the 8 people are: A, a, B, b, C, c, D, d

Using your approach, one possible outcome is as follows:

Stage 1: select B, Stage 2: select d, Stage 3: select A

Here's another possible outcome:

Stage 1: select B, Stage 2: select A, Stage 3: select d

Another possible outcome:

Stage 1: select d, Stage 2: select A, Stage 3: select B

Another possible outcome:

Stage 1: select d, Stage 2: select B, Stage 3: select A

Another possible outcome:

Stage 1: select A, Stage 2: select B, Stage 3: select d

Another possible outcome:

Stage 1: select A, Stage 2: select d, Stage 3: select B

Your approach treats the 6 outcomes above as 6 different outcomes, when in actuality, they are all THE SAME.

Since you have counted every outcome 6 times, we need to divide your final answer by 6.

This will give you the correct answer.

Cheers,

Brent

## Hi Brent,

Thank you for your explanation. Just a small question since I tried to answer it in another way. If I had to calculate the number of ways in which the two twins are from same group, how would I have gone about doing so?

Thanks

## Here's my full solution that

Here's my full solution that follows that approach: https://www.beatthegmat.com/combinations-t123249.html#828080

Cheers,

Brent

## https://gmatclub.com/forum

Why do we use different methods for rectangle and triangles. I calculated it as 98*8 for X and 11*1

0 for Y just as I did for triangle. In both cases X & Y take only two values.

## Question link: https:/

Question link: https://gmatclub.com/forum/rectangle-abcd-is-constructed-in-the-coordina...

I'm assuming you're referring to RIGHT triangles, yes?

The big difference is that triangles have 3 sides and rectangles have 4 sides. So, this will likely affect the outcomes.

Notice that, if I tell you that two diagonally opposite points in a RECTANGLE are (1,3) and (10,7), then I automatically know that the other two points in the rectangle are (1,7) and (10,3)

Conversely, if I tell you that two endpoints of the hypotenuse of a right TRIANGLE are (1,3) and (10,7), then there are TWO possible locations for the 3rd point of the triangle. The 3rd point can be EITHER at (10,3) OR at (1,7)

Does that help?

Cheers,

Brent

## Ohkay! So for a rectangle, we

## That's correct.

That's correct.

## https://gmatclub.com/forum

In this question - if we use combinations and choose x=3,4 and y = 4,5, dont we get a square instead of a rectangle? Please advise

## Question link: https:/

Question link: https://gmatclub.com/forum/rectangle-abcd-is-constructed-in-the-coordina...

A rectangle is any quadrilateral that has 4 right angles.

As such, a square is a certain type of rectangle (one in which all 4 sides have equal length)

For more on this, watch: https://www.gmatprepnow.com/module/gmat-geometry/video/874

Cheers,

Brent

## Hi Brent,

https://www.beatthegmat.com/a-family-consisting-of-one-mother-t103048.html

My approach to this question was as follows -

Stage 1 : the driver's seat- can be accomplished in 2 ways

Stage 2 : Backseat middle seat - can be accomplished in 2, as none of the sisters can't sit together and 1 parent already seated at the driver's seat

Stage 3 : passenger's seat front - can be accomplished in 4 ways

Stage 4 : 2 remaining backseats - can be in 2 and 1 ways

Total = 2x2x4x2x1 = 32 ways

What do you think?

Thanks in advance!

## Question link: https://www

Question link: https://www.beatthegmat.com/a-family-consisting-of-one-mother-t103048.html

Great idea to use the middle backseat to keep the sisters separated!

Cheers,

Brent

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