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Perumtation Problem - need help

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by Night reader » Fri Nov 05, 2010 4:23 pm
rtaha2412 wrote:Ann, Mark, Dave and Paula line up at a ticket window.. In how many ways can they arrange themselves so that Dave is third in line from the window?

a 24
b 12
c 9
d 6
e 3
There are 12 possible ways of arranging for Dave in the third place from the window.

This is a typical permutation problem and let's arrange four spots in our line

#1__ #2__ #3__ #4__ we don't know how line is vectored, i.e. in the direction or opposite to the window; so we should assume our spots #2 and 3 are both positioned in the third spot in the line.

For the first spot we have 3 arrangements, for the second and the third spots we have got 2 arrangements, for last the fourth spot we have 1 arrangement. Calculation of the possible arrangements in a permutation above gives us 3*2*2*1=12.

Pick answer (b).
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by Taniuca » Fri Nov 05, 2010 4:28 pm
POSITIONS
1 2 3 4
A B D P

The answer will be 3! = 6 possible options to place A,M & P in the 1 2 and 4 position.

how to:

A B D P
A P D B
B A D P
B P D A
P A D B
P B D A
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by Night reader » Fri Nov 05, 2010 5:28 pm
rtaha2412 wrote:Ann, Mark, Dave and Paula line up at a ticket window.. In how many ways can they arrange themselves so that Dave is third in line from the window?

a 24
b 12
c 9
d 6
e 3
rtaha, what's oa?
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by bkk_marc » Sun Nov 07, 2010 12:38 am
I think the answer would be 6.

Just like how Taniuca explained.

I don't think it would be 12 because you must assume that the window is on one side and do the calculations from there.

If you are adding both possibilities of the windows being on either side, then you are doubling the results. I could be wrong, but I would also answer 6.

Do you have the answer?
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by ikaplan » Sun Nov 07, 2010 2:34 am
There are 4 places in the queue:

#1___ #2___ #3___ #4___

according to the problem, Dave should be third from the ticket window

window: #1___ #2___ #3Dave #4___

for the first person there are 3 arrangement choices, for the second one 2 choices and for the thrid one- only one choice

therefore, there are 3! possible arrangements

In my opinion, D (6) is the correct answer
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by Geva@EconomistGMAT » Sun Nov 07, 2010 5:26 am
ikaplan wrote:There are 4 places in the queue:

#1___ #2___ #3___ #4___

according to the problem, Dave should be third from the ticket window

window: #1___ #2___ #3Dave #4___

for the first person there are 3 arrangement choices, for the second one 2 choices and for the thrid one- only one choice

therefore, there are 3! possible arrangements

In my opinion, D (6) is the correct answer
Completely agree. The general principle in these questions is to break down the problem into step by step (as ikaplan demonstrated above), then start where you are limited. If David must be in the third place, then we only have one option in that place: David. After that, ask yourself, one step at a a time: How many options do I have for the first place? for the second? for the last remaining?

Let's mix it up a bit. Same question, only now David can stand anywhere (not necessarily third), as long as he's next to Ann. How many arrangements are there?
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by georgeanand » Sun Nov 07, 2010 8:18 am
rtaha2412 wrote:Ann, Mark, Dave and Paula line up at a ticket window.. In how many ways can they arrange themselves so that Dave is third in line from the window?

a 24 b 12 c 9 d 6 e 3
Third position can be filled one way.
The other positions can be filled 3 x 2 x 1 ways - hence 6 ways.
The arrangement can be left to right or right to left. hence there 6 +6 ways= 12 ways
Further modifications lets forget that dave sits in 3rd place
If two people have to sit always together club them as one entity. Hence arrangement = 2X 3!
If two peolpe shouldnt be sitting near each other,arrangement =4!- 2x3!.
There are many more modifications which can be framed

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by rkanthilal » Sun Nov 07, 2010 1:04 pm
Geva@MasterGMAT wrote:
Let's mix it up a bit. Same question, only now David can stand anywhere (not necessarily third), as long as he's next to Ann. How many arrangements are there?
Hi Geva,

I'm getting 12 for your question. If we group Ann and David together we have three "objects" to arrange.
This gives us 3!=6 arrangements with Ann before David and another 3!=6 arrangements with David before Ann for a total 12.

Now, if the restriction was that David and Ann cannot stand next to each other, then would we just subtract these 12 arrangements form the total possible ways to arrange four people?

So,
Total arrangements (no restrictions) = 4! = 24
Arrangements with Ann next to David = (3!)2 = 12
Arrangements with Ann and David Separated = 24 - 12 = 12

Is this correct?

Thanks
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by Geva@EconomistGMAT » Mon Nov 08, 2010 12:02 am
rkanthilal wrote:
Geva@MasterGMAT wrote:
Let's mix it up a bit. Same question, only now David can stand anywhere (not necessarily third), as long as he's next to Ann. How many arrangements are there?
Hi Geva,

I'm getting 12 for your question. If we group Ann and David together we have three "objects" to arrange.
This gives us 3!=6 arrangements with Ann before David and another 3!=6 arrangements with David before Ann for a total 12.

Now, if the restriction was that David and Ann cannot stand next to each other, then would we just subtract these 12 arrangements form the total possible ways to arrange four people?

So,
Total arrangements (no restrictions) = 4! = 24
Arrangements with Ann next to David = (3!)2 = 12
Arrangements with Ann and David Separated = 24 - 12 = 12

Is this correct?

Thanks
To the letter :).

When several things need to be lumped together, lump them as a single "entity" and figure out the number of arrangements of the "new" number of entities, then think about the internal order of the "double entity" itself and multiply the number of permutations accordingly.
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by sana.noor » Tue Jul 30, 2013 10:50 am
it should be 6 not 12....experts am i right?
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by Matt@VeritasPrep » Tue Jul 30, 2013 1:22 pm
sana.noor wrote:it should be 6 not 12....experts am i right?
As given in the first post in the thread (Dave is 3rd), the answer is 6.

Given the extra condition (mentioned later in the thread) that Dave is next to Ann, the answer is 12.
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