Hi,
could someone list (or link to a similar thread) all necessary formulas for the GMAT? With examples?
e.g. n! , (n-k)!, (n k) and so on ?
Which combination/permutation formulas are needed for GMAT??
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Please find attached the required concepts with examples.
I hope it helps.
P.S. I personally don't recommend too much formula oriented techniques to solve the questions as it leads to confusion and incapacity to solve various types of surprising questions.
I hope it helps.
P.S. I personally don't recommend too much formula oriented techniques to solve the questions as it leads to confusion and incapacity to solve various types of surprising questions.
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So you suggest, just "thinking" is the way to go?
I partly agree with you, but on the other hand knowing formulas saves time?
I partly agree with you, but on the other hand knowing formulas saves time?
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Hi confused13,
For permutation and combination questions, you have some options when it comes to HOW you answer the question. Depending on the complexity of the situation and the "size" of the answer choices, you might need to do a lot of math OR you might be able to just "brute force" the situation.
For example, if you have 5 different markers and you want to figure out how many different combinations of 2 there are, how would you go about solve (order does not matter)? You could use the combination formula OR you could just write down all of the combinations. 8-year olds can solve this problem with crayons by just drawing it out. If you call the markers A, B, C, D and E, then you can name all the options relatively quickly.
In contrast, if it's 10 markers and you want to figure out how many different combinations of 4 there are, then you'd probably want to use the Combination Formula:
N!/[K!(N-K)!]
To that end, knowing the Combination Formula is a MUST.
Personally, I've never used the "formal" Permutation Formula to solve a GMAT question, but it doesn't bother me if someone else wants to. GMAT questions usually can be solved in a couple of different ways, so how you choose to do so is up to you.
GMAT assassins aren't born, they're made,
Rich
For permutation and combination questions, you have some options when it comes to HOW you answer the question. Depending on the complexity of the situation and the "size" of the answer choices, you might need to do a lot of math OR you might be able to just "brute force" the situation.
For example, if you have 5 different markers and you want to figure out how many different combinations of 2 there are, how would you go about solve (order does not matter)? You could use the combination formula OR you could just write down all of the combinations. 8-year olds can solve this problem with crayons by just drawing it out. If you call the markers A, B, C, D and E, then you can name all the options relatively quickly.
In contrast, if it's 10 markers and you want to figure out how many different combinations of 4 there are, then you'd probably want to use the Combination Formula:
N!/[K!(N-K)!]
To that end, knowing the Combination Formula is a MUST.
Personally, I've never used the "formal" Permutation Formula to solve a GMAT question, but it doesn't bother me if someone else wants to. GMAT questions usually can be solved in a couple of different ways, so how you choose to do so is up to you.
GMAT assassins aren't born, they're made,
Rich
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As Rich points out, knowing how to calculate combinations is a MUST.
One option is to use the combination formula: N!/[K!(N-K)!]
However, most combinations on the GMAT can be calculated quickly in our heads.
If anyone is interested, we have a free video that explains how to do that: https://www.gmatprepnow.com/module/gmat-counting?id=789
Cheers,
Brent
One option is to use the combination formula: N!/[K!(N-K)!]
However, most combinations on the GMAT can be calculated quickly in our heads.
If anyone is interested, we have a free video that explains how to do that: https://www.gmatprepnow.com/module/gmat-counting?id=789
Cheers,
Brent
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I'd also like to point out that most counting questions on the GMAT can be solved by applying the Fundamental Counting Principle (FCP). The FCP is easy to apply, and typically does not require any formulas. If a formula is required, it is almost always the combinations formula.
Here's an article I wrote on this topic: https://www.beatthegmat.com/mba/2013/07/ ... ons-part-i
For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775
Then you can try solving the following questions:
EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html
- https://www.beatthegmat.com/mouse-pellets-t274303.html
MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html
DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/ps-counting-t273659.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/please-solve ... 71499.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/laniera-s-co ... 15764.html
Cheers,
Brent
Here's an article I wrote on this topic: https://www.beatthegmat.com/mba/2013/07/ ... ons-part-i
For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775
Then you can try solving the following questions:
EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html
- https://www.beatthegmat.com/mouse-pellets-t274303.html
MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html
DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/ps-counting-t273659.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/please-solve ... 71499.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/laniera-s-co ... 15764.html
Cheers,
Brent
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Yes, 100% memorize the permutations and combinations formula. Most permutations will NOT require the formula, and most combinations WILL (esp. when combined with probability).
I would also suggest you memorize how to deal with circular permutations.
For example, there are 7 people and a round table with 5 seats. How many arrangements are possible?
So first we're wondering, how many ways to choose 5 from 7? This is a simple Comb:
7C5 = 7! / 5!2! = 7 x 6 / 2 = 42/2 = 21 ways
So now for each of those ways, we're wondering, how many ways can we order 5 people around a table?
For any table with "x" seats, the number of possible arrangements is (x-1)!, so here 4! = 4 x 3 x 2 = 24.
21 x 24 = 504
These are really advanced concepts, so I wouldn't kill yourself on them unless you're looking for a darn-near perfect Quant score.
Good luck!
I would also suggest you memorize how to deal with circular permutations.
For example, there are 7 people and a round table with 5 seats. How many arrangements are possible?
So first we're wondering, how many ways to choose 5 from 7? This is a simple Comb:
7C5 = 7! / 5!2! = 7 x 6 / 2 = 42/2 = 21 ways
So now for each of those ways, we're wondering, how many ways can we order 5 people around a table?
For any table with "x" seats, the number of possible arrangements is (x-1)!, so here 4! = 4 x 3 x 2 = 24.
21 x 24 = 504
These are really advanced concepts, so I wouldn't kill yourself on them unless you're looking for a darn-near perfect Quant score.
Good luck!
Vivian Kerr
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