# Lesson: Fundamental Counting Principle Example

## Comment on Fundamental Counting Principle Example

### Hello Brent,

Hello Brent,
For the last question of the reinforcement activities (http://www.beatthegmat.com/permutation-gmat-t287098.html),
you write that after the first step of choosing a digit from 9 possibilities,you have to chose the position of the second digit. I quiet don't understand why since we can simply say that the second one must be equal to the first and then choose from 8 possibilities left for the last digit. Could you explain this stage please ?
Thank you ,
Ben ### Glad to help, Ben. You're

Glad to help, Ben. You're referring to this question: http://www.beatthegmat.com/permutation-gmat-t287098.html

Your strategy would work if it were the case that the 1st and 2nd digits are the same, but we must count ALL CASES where two of the digits are the same.

Some examples include: 311, 585, 668, etc.

So, two digits are the same, and the other digit is different.

In my solution, we first examine the 1 digit that is different.

First, we determine what number that digit is (stage 1)

Second, we determine where that digit is located (stage 2)

Third, we select the value of the two digits (which are the same (stage 3)

Does that help?

### Dear Brent,

Dear Brent,

With reference to
*GMAT practice question (difficulty level: 500 to 650) - Beat the GMAT*

Would it be correct to solve it using a method used in one of the earlier questions from the last video.

We'll first arrange the model groups according to continents.
Since there are 3 continents, there are 3! ways of arranging them
= 6.

Now for each of the continents we can arrange the models to stand on either side of each other meaning now we have
2 x 2 x 2 = 8.

Therefore, the total number of ways to arrange the models
= 8 x 6 = 48. That's a perfectly valid approach - great work! ### Hi Brent,

Hi Brent,

Is this approach correct?

The ways the girls can be seated in the end chairs: 2! ways

G _ _ _ G

The ways guys can be seated in the chairs in between,

G (B B B) G is 3! ways

By fundamental counting principle we get 2!*3!= 12 ways ### That approach works perfectly

That approach works perfectly - nice work!

Cheers,
Brent

### https://gmatclub.com/forum

https://gmatclub.com/forum/there-are-six-different-models-that-are-to-appear-in-a-fashion-show-200514.html

Hey,
This can be solved as 6*1*4*1*2*1 right? ### That looks correct.

That looks correct.
Here's my full solution to that question: https://gmatclub.com/forum/there-are-six-different-models-that-are-to-ap...