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Comment on Fundamental Counting Principle Example
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 ,
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?
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
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.
Question link: http://www
Question link: http://www.beatthegmat.com/seating-arrangement-t222030.html
That's a perfectly valid approach - great work!
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!
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...
Hi Brent, How can this
Team A and Team B are competing against each other in a game of tug-of-war. Team A, consisting of 3 males and 3 females, decides to line up male, female, male, female, male, female. The lineup that Team A chooses will be one of how many different possible lineups?
Here's my full response:
Here's my full response: https://gmatclub.com/forum/team-a-and-team-b-are-competing-against-each-...
I just want to clarify the information - "the girls must sit on the end chairs" that means the #1 seat and #5 occupied by girls?
At first time I read it, I thought just only #5 occupied by girls.
That's right, "the girls must
That's right, "the girls must sit on the end chairs" = "the girls occupy seats #1 and #5"
Hey Brent, So this one Im
Good observation. I strayed
Good observation. I strayed from my usual approach in order to answer the question in one step.
The alternative is to recognize that there are three different ways to satisfy the given information.
I use that approach here: https://gmatclub.com/forum/of-the-three-digit-positive-integers-that-hav...
You'll find that this solution is similar to the way I typically solve questions :-)
love it! thank you! Of course