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At an election meeting 10 speakers are to address the

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New post 04 Oct 2009, 04:19
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At an election meeting 10 speakers are to address the meeting. The only protocol to be observed is that whenever they speak the pm should speak before the mp and the mp should speak before the mla.In how many ways can the meeting be addressed?
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Re: permutation  [#permalink]

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New post 04 Oct 2009, 10:25
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appuamar wrote:
tnx a lot guys..only de ans was given so din understand prop


Look at the problem this way: 10 speakers can be arranged in 10! ways, hope this line is clear. Now, in all these ways three persons (mp, pm and mla: let's call them 1,2 and 3) can be arranged in 3!=6 ways:
123
132
213
231
312
321

From all these combinations only one (123) is good for us. As these 6 possibilities will be evenly distributed in 10! ways (meaning that all 6 combinations will have the same number of ways, as combination doesn't favor any of them), so our needed combination (123) will be 1/6 of all possible combinations -->10!/6

Hope now it's clear.
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Re: permutation  [#permalink]

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New post 04 Oct 2009, 07:34
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Hi appuamar,

Your post was a little hard to read--may I suggest cutting and pasting problems, rather than paraphrasing?--but I think I've got a solution.

The PM, MP, and MLA always speak in that order, though at possibly spread out among the ten positions. the means there are 10C3 ways to place those three speakers (i.e. PM 1st, MP 2nd, MLA 3rd; PM 1st, MP 2nd, MLA 4th...)

10C3 = 10! / 3!(10-3)! = 10*9*8 / 3*2 = 120. There are 120 ways to arrange the PM, MP, and MLA.

Now, I can't tell from your transcription of the problem as to whether this is the answer to the question, or whether it is looking for the ways to arrange all 10 speakers. If we need to arrange all 10, then the remaining 7 speakers can be ordered in 7! = 7*6*5*4*3*2 = 5040 ways. However, as the final result ends up being over 600,000, it doesn't strike me as a very GMAT-like answer.

FInally, on the GMAT, there is a strong possibility that we could eliminate trap answers or use common sense to make this problem much less math-intensive. What were the answer choices?
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Re: permutation  [#permalink]

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New post 04 Oct 2009, 08:57
Firstly m so sry i tried cuttin it n pastin it here but it wasen possible so had to type the whole thing out. wel the answer is 10!/6.The question asks for ways in which ten people can address the meeting the only constraint being PM needs to precede MP and the MP needs to precede the MLA. I hope ma explanation is pretty clear.
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Re: permutation  [#permalink]

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New post 04 Oct 2009, 09:39
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OKay, given that the questions was asking for the total number of addresses, 10!/6 actually works fine. There are 10! ways to arrange 10 people. However, we must have PM/MP/MLA. We cannot have any other combination, such as MP/PM/MLA or MLA/PM/MP. Therefore, since there are 6 combinations of those three people (3!) of which only 1 is acceptable, only 1/6 of the total combinations are valid--that is to say, 10!/6.
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New post 04 Oct 2009, 09:44
Bunuel wrote:
10 speakers can be arranged in 10! ways. Protocol to be observed only one possibility from 3! is appropriate, so, total number of ways=10!/3!=10!/6


Let's try to state the problem in another way:

At an election meeting 10 speakers are to address the meeting.The only protocol to be observed is that whenever they speak the pm should speak right before the mp and the mp should speak right before the mla.In how many ways can the meeting be addressed? (The answer would be different)
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Re: permutation  [#permalink]

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New post 04 Oct 2009, 10:07
tnx a lot guys..only de ans was given so din understand prop
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Re: permutation  [#permalink]

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New post 05 Oct 2009, 01:11
10!/6 (or 10!/3!) works if those three have to give speech just one after another (There are no other speakers among those three). Otherwise, it should be 10!/10c3. Its not quite clear for me from the question.
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New post 05 Oct 2009, 06:48
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eresh wrote:
10!/6 (or 10!/3!) works if those three have to give speech just one after another (There are no other speakers among those three). Otherwise, it should be 10!/10c3. Its not quite clear for me from the question.


Not so: "if those three have to give speech just one after another (There are no other speakers among those three)", the formula would be different.

In fact this is the Q I asked in my previous post:

"Let's try to state the problem in another way:

At an election meeting 10 speakers are to address the meeting.The only protocol to be observed is that whenever they speak the pm should speak right before the mp and the mp should speak right before the mla.In how many ways can the meeting be addressed? (The answer would be different)"

In this case the answer would be: 8!, because 3 speakers are fixed in one order, plus 7 other speakers, gives us total 8 items to be arranged=8!.

In the initial case between 3 (PM, MP and MLA) we could have other speakers.

Total number of arrangement 10!. PM, MP and MLA arrangement 3!, from which only one "any number of other speaker(s) - PM - any number of other speaker(s) -MP - any number of other speaker(s) - MLA - any number of other speaker(s)" is meeting the protocol.

So, only in 1 of 6 cases protocol would be observed and in other 5 not, which gives us 10!/6 possible arrangements with protocol observed.
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New post 17 Oct 2009, 15:24
I get so confused with this type of probability questions :shock:

Just to clear my concept......if the question was-

At an election meeting 10 speakers are to address the meeting.The only protocol to be observed is that whenever they speak the pm, the mp and the mla should speak one after another (in no particular order).In how many ways can the meeting be addressed?"

In this case, would it be 8!*3! ?
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New post 17 Oct 2009, 19:36
eresh wrote:
I get so confused with this type of probability questions :shock:

Just to clear my concept......if the question was-

At an election meeting 10 speakers are to address the meeting.The only protocol to be observed is that whenever they speak the pm, the mp and the mla should speak one after another (in no particular order).In how many ways can the meeting be addressed?"

In this case, would it be 8!*3! ?


If I understood correctly you are offering third case: 3 fixed but no matter what order.

If so, yes it would be 8!3!.

Let see it on similar case but with smaller numbers 5 speakers: 3 to be fixed, order doesn't matter and fourth and fifth, lets call them A and B:

Three cases:
1. A, B (B,A) and after them our 3 speakers=2!3!
2. 3 speakers and then AB(BA)=2!3! again.
3. A, 3 speakers, B OR B, 3 speakers, A=2!3! again.
So total 3*2!*3!=3!*3!

Another way: fix 3 as 1 speaker, plus A and B=3, ways between them 3! and we should multiply this by the ways 3 our speakers can arrange between each other=3!. Total=3!*3! The same.

Correct.

Hope it's clear.
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New post 18 Oct 2009, 00:27
Thank you. It is clear indeed.
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