Arrangements

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Arrangements

by 4meonly » Fri Nov 14, 2008 6:02 am
Tom, Jerry and Donald and other three people sit in a line. From left ot right, if Tom cannot sit on the first seat, Jerry cannot sit on the second seat and Donald cannot sit on the fourth seat, how many different arrengements are possible?


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by cramya » Fri Nov 14, 2008 7:59 am
A good one! Cant get to the OA though :x

I am missing something here with this approach.. Thoughts/Suggestions welcome....

Lets consider arrangements wihtout any constraints = 6*5*4*3*2*1 = 720

Lets find all the arrangemnts that would offend the restirction

Tom alone in the first place with Jerry and Donald not in 2nd and 4th

T 4 4 2 2 1 = 64

Jerry alone in 2nd place wiht Tom not in 1st and Donald not in 4th

4 J 4 2 2 1 = 64

D alone in 4th with Tom not in 1 and Jerry not in 2

4 3 3 D 2 1 = 72

Tom in 1 Jerry in 2 Donaled not in 4th

T J 4 2 2 1 =16

Tom in 1 Donald in 4th and Jerry not in 2nd

T 3 3 D 2 1 = 18

Jerry in 2 Donald in 4 and Tom not in 1

3 J 3 D 2 1 = 18

Tom in 1 Jerry in 2 and Donald in 4

T J 3 D 2 1 = 6


I am geting 462 whnd I add these making the no of arrangments 720-258 = 462

What arrangement that need tO be exlcuded to get to the OA, am I missing??

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by logitech » Fri Nov 14, 2008 10:17 am
cramya wrote:A good one! Cant get to the OA though :x

I am missing something here with this approach.. Thoughts/Suggestions welcome....

Lets consider arrangements wihtout any constraints = 6*5*4*3*2*1 = 720

Lets find all the arrangemnts that would offend the restirction

Tom alone in the first place with Jerry and Donald not in 2nd and 4th

T 4 4 2 2 1 = 64

Jerry alone in 2nd place wiht Tom not in 1st and Donald not in 4th

4 J 4 2 2 1 = 64

D alone in 4th with Tom not in 1 and Jerry not in 2

4 3 3 D 2 1 = 72

Tom in 1 Jerry in 2 Donaled not in 4th

T J 4 2 2 1 =16

Tom in 1 Donald in 4th and Jerry not in 2nd

T 3 3 D 2 1 = 18

Jerry in 2 Donald in 4 and Tom not in 1

3 J 3 D 2 1 = 18

Tom in 1 Jerry in 2 and Donald in 4

T J 3 D 2 1 = 6


I am geting 462 whnd I add these making the no of arrangments 720-258 = 462

What arrangement that need tO be exlcuded to get to the OA, am I missing??
Cramya, I did not see anything you were missing in your calculation. But there should be a GMAT solution to this question. And I really don't trust all of the OA's.
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by mental » Sat Nov 15, 2008 5:51 am
I dont know for sure, but they should be equal as the constraints are same!!
Tom alone in the first place with Jerry and Donald not in 2nd and 4th
T 4 4 2 2 1 = 64

Jerry alone in 2nd place wiht Tom not in 1st and Donald not in 4th
4 J 4 2 2 1 = 64

D alone in 4th with Tom not in 1 and Jerry not in 2
4 3 3 D 2 1 = 72
and
Tom in 1 Jerry in 2 Donaled not in 4th
T J 4 2 2 1 =16

Tom in 1 Donald in 4th and Jerry not in 2nd
T 3 3 D 2 1 = 18

Jerry in 2 Donald in 4 and Tom not in 1
3 J 3 D 2 1 = 18
first three cases should be 72 each
T 4 3 3 2 1...........2nd place by 4, 4th place by 3, 3rd by 3, 5th by 2, 6th 1

similarly next three cases should be 18
T J 3 3 2 1...........4th place by 3, 3rd place by 3, 5th by 2, 6th by 1

720 - (72 + 72 + 72 + 18 + 18 + 18 +6) = 444

STILL NOT SAME AS OA!!!!!

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by cramya » Sat Nov 15, 2008 9:40 am
I dont know for sure, but they should be equal as the constraints are same!!

I am not sure on this I have seen same constraints applied to different positions in the line up change the overall possible arrangements amongst them.

However u could be correct also. Lets wait to hear back from few others also...

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by stop@800 » Sat Nov 15, 2008 11:46 am
My attempt:

Please paste the official explanation also.


Total no of possible arrangements = 6! = 720

Part: 1
Arrangements when T is at first place = 5! [All these cases needs t be subtracted from 6!]

Part: 2
Total Arrangements when S is at second place = 5!
Nos this 5! also includes cases when T is at first place [remember these cases have already been subtracted in part 1]
so arrangements when S is at second place and T is at first place = 4!
Hence
Total Arrangements when S is at second place, which have still not been considered = 5! - 4!

Part: 3
This is a bit tricky one
Total Arrangements when D is at fourth place = 5!
Now we need to subtract cases which has already been considered in part 1 and part 2.

S fixed at fourthe place
Arrangements when T was at first place: 4!
Arrangements when J was at second place: 4!
so
Arrangements when D is at fourth place = 5! - 2*4!
Just wait
The 4! (Arrangements when T was at first place) also includes the case when J is at seond place.

similarly

The 4! (Arrangements when J was at second place) also includes the case when T is at first place.

hence
we have counted these cases twice
btw, how many such arrangements are there, its 3!

so finally we have
5!
subtract J and T part
5! - 2*4!
Now we need to add 3! [as this has been subtracted twice]
5! - 2*4! + 3!


Desired Total arrangements = 6! - (5!) - (5!-4!) - (5!-2*4!+3!)

I know its a bit tricky, please feel free if you need detailed explanation.

4meonly, Whats the source of this excellent Qn??

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by cramya » Sat Nov 15, 2008 12:17 pm
Stop@800 thats good.

Where do u see the problem in my approach?(i.e what need to be added/modified)

I have asked a few other experts on this forum to look at too.

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by stop@800 » Sat Nov 15, 2008 12:47 pm
cramya wrote: Tom alone in the first place with Jerry and Donald not in 2nd and 4th

T 4 4 2 2 1 = 64
Lets take the second position
We have taken it as 4 cases, because we are ignoring Jerry.

There is a possibility that Donald will come at this place and when considering Donald's place you have one extra choice as Donald is not available.

Hope this helps!!!

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by lunarpower » Wed Nov 19, 2008 3:31 am
i think this problem is a bit too long and involved / complicated to show up on the real test, but i've seen some surprises, so you never know.

here's the way i would do this: i would count up all the disallowed arrangements of just tom, jerry, donald, and then subtract them from the total number of arrangements of those people (which is 6 x 5 x 4, or 120).

here are the disallowed arrangements, given in the order tom, jerry, donald:
123 124 125 126 134 135 136 145 146 156
132 142 152 162 143 153 163 154 164 165 (notice the symmetry between this row and the last row)
that's all the arrangements in which tom is in #1. now we have to list the ones in which jerry is in #2, but tom isn't in #1 (because we already have the ones where tom is in #1):
321 324 325 326 421 423 425 426 521 523 524 526 621 623 624 625
finally, we need the arrangements in which donald is in #4, but tom isn't in #1 and jerry isn't in #2:
214 234 254 264 314 354 364 514 534 564 614 634 654
that's 49 disallowed arrangements total.
therefore, there are 120 - 49 = 71 allowed arrangements.

for each of those 71 allowed arrangements of tom, jerry, and donald, there are 3! = 6 ways to arrange the remaining 3 individuals. therefore, the total # of arrangements is 71 x 6 = 426.

again, as i said before, this problem is a bit much. i could see something like it showing up on the test, though, if it were pared down to 5 individuals (a minor change, but one that would approximately halve the amount of necessary computation).
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by cramya » Wed Nov 19, 2008 6:21 am
Ron,
Could you please point me to what I am missing in my arrangements/approach?

Regards,
Cramya

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by lunarpower » Wed Nov 19, 2008 3:32 pm
cramya wrote:Ron,
Could you please point me to what I am missing in my arrangements/approach?

Regards,
Cramya
you've basically made the same single mistake several times. here's the basic idea: if you're going to multiply possibilities, then those numbers of possibilities have to be independent of one another (i.e., the numbers have to work out the same way regardless of who is chosen).

here's one place where you ran afoul of that principle:
Tom in 1 Jerry in 2 Donaled not in 4th

T J 4 2 2 1 =16
wrong.

you can say there are four choices for the third spot, but then you run into trouble: you don't know how many choices there are for the fourth spot.
here's the problem: if you happened to pick donald for the 3rd spot, then there will be 3 choices for the 4th spot (everyone who's left; no need to worry about donald anymore). on the other hand, if you pick anyone but donald for the 3rd spot, then the numbers are 2, 2, 1, as you wrote.

this was articulated by stop800 in one of the above posts.

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here's further proof that this method is not working:
Tom in 1 Donald in 4th and Jerry not in 2nd

T 3 3 D 2 1 = 18
now this is actually a correct answer, because (by accident, it seems) you're choosing the restricted slot (i.e., the second slot) first.
but the reason you should notice there's a problem is that you have two IDENTICAL situations - two people in fixed locations, a third person with one prohibited location, and three people who can go anywhere - and yet you get two different numbers of possibilities (16 vs. 18). this is of course impossible: if you present two exactly identical situations, you have to get the same number of possibilities both times.

this observation won't tell you what to do to fix the problem, nor will it even diagnose the problem, but at least it's a symptom.

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by the way, you can fix the '16' above by picking the 4th slot (the restricted one) first:

t j 3(chosen second) 3(chosen first) 2 1 = 18.
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by lunarpower » Wed Nov 19, 2008 3:37 pm
cramya wrote:
I am not sure on this I have seen same constraints applied to different positions in the line up change the overall possible arrangements amongst them.
impossible, unless the positions in the lineup are somehow distinguishable from each other. in this problem, they aren't; there are just six spots numbered #1 through #6, none of which has any distinguishing characteristics.

in math problems, symmetry always begets symmetry. it can't work out any other way.
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by lunarpower » Wed Nov 19, 2008 3:44 pm
cramya wrote:Tom alone in the first place with Jerry and Donald not in 2nd and 4th

T 4 4 2 2 1 = 64

Jerry alone in 2nd place wiht Tom not in 1st and Donald not in 4th

4 J 4 2 2 1 = 64

D alone in 4th with Tom not in 1 and Jerry not in 2

4 3 3 D 2 1 = 72
i don't really see any simple way in which you can do these by multiplication, because there's no simple order in which you can choose the restricted spots first.

let's attack the first one: tom in the first place, jerry NOT in the second place, donald NOT in the fourth place.

here's the way i would do this:

* the TOTAL number of arrangements in which tom is in place #1 is 5!, or 120.

* the total number of arrangements that are NOT allowed is the number of arrangements in which either jerry is in spot #2, or donald is in spot #4, or both. for this, we can use the classic "either/or formula": (# of arrangements with jerry in 2nd) + (# of arrangements with donald in 4th) - (# of arrangements with both).
that's (4!) + (4!) - (3!), or 42.

* subtract to find the number of arrangements you want: 120 - 42 = 78.

--

if you look at the post by user "mental", that user has 72 instead of 78 for each of these first three situations. that's a difference of six, which means that there are 3 x 6 = 18 possibilities left unaccounted for by "mental" (because there are three such situations).
given that mental got 444, which is exactly 18 more than the official answer, the fixes above will rectify the solution.

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i will repeat my previous assertion that this problem is too long and involved to appear on the gmat, although components of its solution would certainly be useful in addressing other problems.
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by lunarpower » Wed Nov 19, 2008 3:46 pm
and by the way, there's nothing wrong with replacing the fourth spot (the spot from which donald is barred) with the third spot.
all that's essential in the problem is that there's some spot in which donald can't go; since none of the spots is "special" in any way, it doesn't matter which spot you disallow for donald.

i say this because i'd have a much easier time keeping things straight if the prohibited spots were #1, #2, #3 respectively than if they were #1, #2, #4. i'm sure most of you would, too.
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by cramya » Wed Nov 19, 2008 7:35 pm
Wow thats going to take quite some time to read and digest the information presented. I am going to though....

Ron, thanks so much for taking your time in explaining it! Much appreciated:-)