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PERMUTATION & COMBINATION

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PERMUTATION & COMBINATION

by ashishjha100 » Thu Jun 19, 2008 10:23 pm
pls help in solving following questions........


1.The no: of ways in which a score of 11 can be made from a throw by 3 persons, each throwing a single die once,is

a.45
b.18
c.27
d.none


2.Between 2 stations A & B there are 12 intermediate stations.The no: of ways in which a train can be made to stop at 4 of these stations so that no two of these halting stations are consecutive is

a.8C4
b.9C4
c.12C4 - 4
d.NONE
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by Ian Stewart » Fri Jun 20, 2008 5:33 am
As is clear at first glance, these can't be real GMAT problems- there are only four answer choices.

The first problem, about summing dice, is normally done by listing the ways you can get the correct sum of 11:

641 (in any of 6 orders)
632 (in any of 6 orders)
551 (in any of 3 orders)
542 (in any of 6 orders)
533 (in any of 3 orders)
443 (in any of 3 orders)

That's 27 ways in total. There's no brilliant short cut here.

I have no idea what the second question even means. GMAT questions are never so badly worded.
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Re: PERMUTATION & COMBINATION

by Motherjane » Fri Jun 20, 2008 6:29 am
ashishjha100 wrote:
2.Between 2 stations A & B there are 12 intermediate stations.The no: of ways in which a train can be made to stop at 4 of these stations so that no two of these halting stations are consecutive is

a.8C4
b.9C4
c.12C4 - 4
d.NONE
I think I can understand the question. Just solved another question the other day and the logic is the same.

The problem is:


If n objects are arranged in a row, then the no: of ways of selecting 3 of these objects so that no two of them are next to each other.

a.(n-3)C3
b.(n-3)C2
c.(n-2)C3
d.none

Solution:
Consider the sum in its simplest form...
let there be 5 objects.Only 2 positions are restricted.Rest 3 picking have to be from remaining 3 ppl..i.e 3C3 ways .

__ X __ X __

For every subsequent addition of object, there is 1 more position to select from.

Hence, if there are n objects..only 2 positions are restricted i.e we can select 3 positions from remaining (n-2) positions.

Hence total selections = (n-2)C3.Ans C.


Taking this logic to the question you have posted, the answer will be

(n-2)C4 = (14-2)C4 = 12C4

So, the answer is (d).
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by Ian Stewart » Fri Jun 20, 2008 9:13 am
Ah, thanks MotherJane for clarifying the language. For whatever reason, I wasn't thinking of the stations as being along the same track from A to B, which made the question seem nonsensical.

So, the question is really asking, if you have 12 things in a row, in how many ways can you choose four of them so that no two are consecutive?

The answer is 9C4. The above argument about 'restrictions' doesn't immediately make intuitive sense to me, but it does give the right answer, so I'm probably not looking at it from the best point of view. You do need to account for having three restricitions, not two, if you are selecting four things. So your formula would become (n-3)C4 = 9C4.

I looked at the problem in two different ways. If you choose any four stations from the first nine, which you can do in 9C4 ways, then insert an extra space between each station, you are certain to get a selection of four from twelve choices where no two are consecutive. And you'll never count the same set twice in this way, which makes the answer 9C4.

I also did this by breaking the problem down into cases. You know there's at least one space between each station (which I'll label S). If we start from a 'word' like:

S_S_S_S

we just need to insert five spaces to get a valid set of four stations from twelve, where no two are consecutive. We can then break down, case by case, how those spaces could be inserted. For example, if we insert five consecutive spaces somewhere, we have five places where we can do it. If we insert four spaces somewhere, and one space somewhere else, we have 5*4 ways to do it. And so on.... the answer again turns out to be 126, or 9C4.

There's probably a generating function approach for this question, and there certainly is for the dice problem above- but that's way, way beyond what's needed for GMAT math. I'll reiterate what I said before: this is not a GMAT question!
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by beeparoo » Sat Jun 21, 2008 5:00 pm
@ashish: What is the OA?
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by parveen110 » Sat Jan 18, 2014 8:06 am
Ian Stewart wrote:Ah, thanks MotherJane for clarifying the language. For whatever reason, I wasn't thinking of the stations as being along the same track from A to B, which made the question seem nonsensical.

So, the question is really asking, if you have 12 things in a row, in how many ways can you choose four of them so that no two are consecutive?

The answer is 9C4. The above argument about 'restrictions' doesn't immediately make intuitive sense to me, but it does give the right answer, so I'm probably not looking at it from the best point of view. You do need to account for having three restricitions, not two, if you are selecting four things. So your formula would become (n-3)C4 = 9C4.

I looked at the problem in two different ways. If you choose any four stations from the first nine, which you can do in 9C4 ways, then insert an extra space between each station, you are certain to get a selection of four from twelve choices where no two are consecutive. And you'll never count the same set twice in this way, which makes the answer 9C4.

I also did this by breaking the problem down into cases. You know there's at least one space between each station (which I'll label S). If we start from a 'word' like:

S_S_S_S

we just need to insert five spaces to get a valid set of four stations from twelve, where no two are consecutive. We can then break down, case by case, how those spaces could be inserted. For example, if we insert five consecutive spaces somewhere, we have five places where we can do it. If we insert four spaces somewhere, and one space somewhere else, we have 5*4 ways to do it. And so on.... the answer again turns out to be 126, or 9C4.

There's probably a generating function approach for this question, and there certainly is for the dice problem above- but that's way, way beyond what's needed for GMAT math. I'll reiterate what I said before: this is not a GMAT question!
Hi Ian,

Let me first tell you that your answers have always been very, very insightful..I find all your solutions presenting totally different, off-book and beautiful approach. Thank you.

This post however is in regard to the second approach of the above problem..Didn't quite get it. Please explain it once more. Thanks.
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by parveen110 » Sat Jan 18, 2014 8:07 am
Ian Stewart wrote:Ah, thanks MotherJane for clarifying the language. For whatever reason, I wasn't thinking of the stations as being along the same track from A to B, which made the question seem nonsensical.

So, the question is really asking, if you have 12 things in a row, in how many ways can you choose four of them so that no two are consecutive?

The answer is 9C4. The above argument about 'restrictions' doesn't immediately make intuitive sense to me, but it does give the right answer, so I'm probably not looking at it from the best point of view. You do need to account for having three restricitions, not two, if you are selecting four things. So your formula would become (n-3)C4 = 9C4.

I looked at the problem in two different ways. If you choose any four stations from the first nine, which you can do in 9C4 ways, then insert an extra space between each station, you are certain to get a selection of four from twelve choices where no two are consecutive. And you'll never count the same set twice in this way, which makes the answer 9C4.

I also did this by breaking the problem down into cases. You know there's at least one space between each station (which I'll label S). If we start from a 'word' like:

S_S_S_S

we just need to insert five spaces to get a valid set of four stations from twelve, where no two are consecutive. We can then break down, case by case, how those spaces could be inserted. For example, if we insert five consecutive spaces somewhere, we have five places where we can do it. If we insert four spaces somewhere, and one space somewhere else, we have 5*4 ways to do it. And so on.... the answer again turns out to be 126, or 9C4.

There's probably a generating function approach for this question, and there certainly is for the dice problem above- but that's way, way beyond what's needed for GMAT math. I'll reiterate what I said before: this is not a GMAT question!
Hi Ian,

Let me first tell you that your answers have always been very, very insightful..I find all your solutions presenting totally different, off-book and beautiful approach. Thank you.

This post however is in regard to the second approach of the above problem..Didn't quite get it. Please explain it once more. Thanks.
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by vipulgoyal » Thu Jan 23, 2014 9:55 pm
|-1-2-3-4-5-6-7-8-9-10-11-12-|

let name these stations with 1 to 12

NO two halting stations are consecutive
means NO two , No three , no four

now use seprator method we can choose any four halting stations(It will not effect combination)

let represent halting station by | and normal station by 0

000|00|0000|0|00

four | can arrange themself bu 4! ways and 8 "0" can be arranged by 8! ways

hence total no ways = 12!/8!*4! d
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