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Question about combination problems

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Question about combination problems  [#permalink]

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New post 23 Feb 2015, 14:55
Good afternoon.

Some combination problems use the equation x!/y!
-ie, how many ways can the letters in COMMON be arranged = 6! / (2!2!)

while others use the equation, n! / (k! (n-k)!)
-ie, how many combinations of grabbing 3 marbles from a bag of 10 marbles = 10! / (3!(10-3)!)

how do you know which one to use?
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Re: Question about combination problems  [#permalink]

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New post 23 Feb 2015, 21:41
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Hi btan219,

The first example that you listed is actually a Permutation question - because you're putting letters in order. The reason why that calculation is divided by 2! and (another) 2! is because there are 2 Ms and 2 Os. If you we were dealing with 6 DIFFERENT letters and forming 6-letter arrangements, then there would be 6! = 720 possibilities. With the word COMMON though....

C(O1)(M1)(M2)(O2)N is the same as...
C(O1)(M2)(M1)(O2)N and a few other variations, but you're not allowed to count them all because they are not unique results. The 2! and 2! remove all of the duplicate entries.

The 'big' question that you're asking is really about 'order' - when reading through the question, you have to ask yourself IF order matters. If it does, then it's a Permutation question; if it does NOT, then it's a Combination question. Thankfully, the questions themselves often offer clues (the word "arrangement" means Permutation, the word "combinations" means Combination).

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Re: Question about combination problems  [#permalink]

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New post 24 Feb 2015, 20:11
btan219 wrote:
Good afternoon.

Some combination problems use the equation x!/y!
-ie, how many ways can the letters in COMMON be arranged = 6! / (2!2!)

while others use the equation, n! / (k! (n-k)!)
-ie, how many combinations of grabbing 3 marbles from a bag of 10 marbles = 10! / (3!(10-3)!)

how do you know which one to use?


hi btan219,
at times the wording of the question does tell you what to use..
1) if the Q asks you to find combinations, selection, order not important, etc you got to use combinations formula...a!/b!(a-b)!...
2) if the Q asks you to find permutations, arrangements, order important, etc you got to use permutations formula...a!/(a-b)!...
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: Question about combination problems  [#permalink]

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New post 26 Feb 2015, 23:14
btan219 wrote:
Good afternoon.

Some combination problems use the equation x!/y!
-ie, how many ways can the letters in COMMON be arranged = 6! / (2!2!)

while others use the equation, n! / (k! (n-k)!)
-ie, how many combinations of grabbing 3 marbles from a bag of 10 marbles = 10! / (3!(10-3)!)

how do you know which one to use?


Check out this post: http://www.veritasprep.com/blog/2011/11 ... binations/
It discusses the difference between the two concepts - permutation and combination - in detail
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New post 27 Feb 2015, 07:39
eh. im still really confused. the veritas confused me even more. i really dont understand what they mean by arranging and not arranging.

1)in the first example, they asked: there are 5 friends but only 3 seats in a row. In how many ways can you make 3 of the 5 friends sit in the 3 seats?
5 combinations can sit in seat 1, 4 combinations can sit in seat 2, and 3 combinations can sit in seat 3.
5x4x3=60
what is the equation being used? i dont see any way a!/b! or a!/(b!(a-b)!) will get 60

2) second example says 5 friends and you have to invite any 3 of them to go with you on a vacation. In how many ways can you do that? they give the answer as 10, they didnt state it but im guess they used 5!/(3!(5-3)!)
but isn't it the same thing as the one above? choosing the first friend, i have 5 to choose from. choosing the second friend i have 4 to choose from. choosing the 3 friend i have 3 to choose from. shouldnt it still be 60?


3) and also, i thought the first problem was a permutation (order matters), so why isn't 5!/3!(5-3)!) used? and isn't the second problem a combination problem, why isn't 5!/3! used?

i know these 2 examples are easy and countable, but i want to be able to know which equations to use, im sure the problems aren't going to be this simple in the gmat. but even though its so simple i really just dont get it at all.

this is really confusing and i feel like if i dont get a basic concept as this, i should just give up. how many problems can you possible get about combinations/permutations on the actual gmat anyways?
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Re: Question about combination problems  [#permalink]

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New post 27 Feb 2015, 08:01
EMPOWERgmatRichC wrote:
Hi btan219,

The first example that you listed is actually a Permutation question - because you're putting letters in order. The reason why that calculation is divided by 2! and (another) 2! is because there are 2 Ms and 2 Os. If you we were dealing with 6 DIFFERENT letters and forming 6-letter arrangements, then there would be 6! = 720 possibilities. With the word COMMON though....

C(O1)(M1)(M2)(O2)N is the same as...
C(O1)(M2)(M1)(O2)N and a few other variations, but you're not allowed to count them all because they are not unique results. The 2! and 2! remove all of the duplicate entries.

The 'big' question that you're asking is really about 'order' - when reading through the question, you have to ask yourself IF order matters. If it does, then it's a Permutation question; if it does NOT, then it's a Combination question. Thankfully, the questions themselves often offer clues (the word "arrangement" means Permutation, the word "combinations" means Combination).

GMAT assassins aren't born, they're made,
Rich


you're saying the COMMON problem is a permutation, but it still uses the a!/b! formula. but i thought permutation are supposed to use a!/(b!(a-b)!)...
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Re: Question about combination problems  [#permalink]

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New post 02 Mar 2015, 00:51
1
btan219 wrote:
eh. im still really confused. the veritas confused me even more. i really dont understand what they mean by arranging and not arranging.

1)in the first example, they asked: there are 5 friends but only 3 seats in a row. In how many ways can you make 3 of the 5 friends sit in the 3 seats?
5 combinations can sit in seat 1, 4 combinations can sit in seat 2, and 3 combinations can sit in seat 3.
5x4x3=60
what is the equation being used? i dont see any way a!/b! or a!/(b!(a-b)!) will get 60

2) second example says 5 friends and you have to invite any 3 of them to go with you on a vacation. In how many ways can you do that? they give the answer as 10, they didnt state it but im guess they used 5!/(3!(5-3)!)
but isn't it the same thing as the one above? choosing the first friend, i have 5 to choose from. choosing the second friend i have 4 to choose from. choosing the 3 friend i have 3 to choose from. shouldnt it still be 60?


3) and also, i thought the first problem was a permutation (order matters), so why isn't 5!/3!(5-3)!) used? and isn't the second problem a combination problem, why isn't 5!/3! used?

i know these 2 examples are easy and countable, but i want to be able to know which equations to use, im sure the problems aren't going to be this simple in the gmat. but even though its so simple i really just dont get it at all.

this is really confusing and i feel like if i dont get a basic concept as this, i should just give up. how many problems can you possible get about combinations/permutations on the actual gmat anyways?


You actually get only 2-3 questions on P&C so don't worry too much.

That said, the basics of P&C that you are struggling with are not that hard to master. Take another example:

There are 5 students:A, B, C, D and E

Case 1: You have to select 3 students to make a quiz team.
Note here that all 3 students have the same position - member of the quiz team. If you select A, B and C, they make the same team as that when you select B, C and A. There is no "first" team member, "second" team member etc. All 3 members have the same position. All you have to do here is SELECT a group of 3 students out of 5. This is done in 5!/3!*2! = 10 ways

Case 2: You have to select 3 students and give them gold, silver and bronze medals.
Now, you SELECT 3 students as before but you also have to ARRANGE them in first second and third positions (gold, silver and bronze). The three students will occupy three different positions. You do have "first" position, "second" position etc here. So selection and arrangement is done in 5!/2! = 60 ways.
This is similar to the case where out of 5 friends, 3 are selected and made to sit in a row because each seat is different. You have a "first" seat, "second" seat and "third" seat so A, B, C is different from B, C, A. Here again, you have to select and arrange so you use 5!/2! = 60 (Mind you, the two formulas are a!/(a-b)! and a!/b!*(a-b)!)

For more details, I suggest you to check out our Combinatorics book. It starts with the basics and goes on to discuss a lot of GMAT relevant concepts.
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Question about combination problems  [#permalink]

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New post 06 Mar 2015, 08:56
yup. no way im gonna get this, ive already read several guides and posts. thanks for the help though guys
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New post 28 Jan 2018, 23:56
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Re: Question about combination problems &nbs [#permalink] 28 Jan 2018, 23:56
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