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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

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