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Count number of ways to arrange 4 people A, B, C, D in a row so that

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tranglenh
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Count number of ways to arrange 4 people A, B, C, D in a row so that [#permalink] New post   10 Aug 2011, 00:45
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Count number of ways to arrange 4 people A, B, C, D in a row so that C, D not sit next to each other

[*] Manhattan solution:
Show Spoiler
(Total number of arrangement - number C, D sit next)
Manhattan approach: pretend that C, D stuck together : then Count the number of ways 2 people not sitting next to each other,
Total number of arrangements: 4! = 24
then number of ways of arrangement so that C, D next to each other is: 3! = 6. Since C, D are distinct, so all number of ways C, D next to each other is: 6x2 = 12.
--> Number of arrangements C, D not sit next: 24 -12 = 12

[*]PFD: (Total number of arrangement - number C, D sit next)
PFD approach: Force C sit on D's right by creating 3 slots for D to sit in: seat 1,2,3. After D choose his seat, S automatically sit on his right.
So, total number of arrangement where C, D next to each other: 3x1 = 3 ( since D can sit on C's right --> then total arrangement is: 3x2 = 6

---> Number of arrangement C, D not sit next: 24 - 6 = 18 (page 83 PFD)
 :roll:
Please help to find the error here? why two answer different


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Re: Count number of ways to arrange 4 people A, B, C, D in a row so that [#permalink] New post   10 Aug 2011, 08:16
tranglenh wrote:
Q: Count number of ways to arrange 4 people A, B, C, D in a row so that C, D not sit next to each other

[*] Manhattan solution: (Total number of arrangement - number C, D sit next)
Manhattan approach: pretend that C, D stuck together : then Count the number of ways 2 people not sitting next to each other,
Total number of arrangements: 4! = 24
then number of ways of arrangement so that C, D next to each other is: 3! = 6. Since C, D are distinct, so all number of ways C, D next to each other is: 6x2 = 12.
--> Number of arrangements C, D not sit next: 24 -12 = 12

[*]PFD: (Total number of arrangement - number C, D sit next)
PFD approach: Force C sit on D's right by creating 3 slots for D to sit in: seat 1,2,3. After D choose his seat, S automatically sit on his right.
So, total number of arrangement where C, D next to each other: 3x1 = 3 ( since D can sit on C's right --> then total arrangement is: 3x2 = 6

---> Number of arrangement C, D not sit next: 24 - 6 = 18 (page 83 PFD)
 :roll:
Please help to find the error here? why two answer different


Firstly when you are forcing D to sit on right of C, you are left with 2 seats hence only 2! ways is possible and not 3!. Secondly You are only assuming C to sit in first place and D on right next to him. C can also take 2nd place, 3rd place and 4th place. So we have

C D _ _ (Here D is forced to right side hence 2 ways)
_ C D _ (Here can sit on right or left of C hence 2*2 = 4 ways)
_ _ C D (Here can sit on right or left of C hence 2*2 = 4 ways)
_ _ D C (Here D is forced to left side hence 2 ways)

Total 12 ways they can sit together. When deducted from 4! we get 24-12 = 12

Hope this clarifies
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Re: Count number of ways to arrange 4 people A, B, C, D in a row so that [#permalink] New post   10 Aug 2011, 08:35
Sudhanshuacharya wrote:
tranglenh wrote:
Q: Count number of ways to arrange 4 people A, B, C, D in a row so that C, D not sit next to each other

[*] Manhattan solution: (Total number of arrangement - number C, D sit next)
Manhattan approach: pretend that C, D stuck together : then Count the number of ways 2 people not sitting next to each other,
Total number of arrangements: 4! = 24
then number of ways of arrangement so that C, D next to each other is: 3! = 6. Since C, D are distinct, so all number of ways C, D next to each other is: 6x2 = 12.
--> Number of arrangements C, D not sit next: 24 -12 = 12

[*]PFD: (Total number of arrangement - number C, D sit next)
PFD approach: Force C sit on D's right by creating 3 slots for D to sit in: seat 1,2,3. After D choose his seat, S automatically sit on his right.
So, total number of arrangement where C, D next to each other: 3x1 = 3 ( since D can sit on C's right --> then total arrangement is: 3x2 = 6

---> Number of arrangement C, D not sit next: 24 - 6 = 18 (page 83 PFD)
 :roll:
Please help to find the error here? why two answer different


Firstly when you are forcing D to sit on right of C, you are left with 2 seats hence only 2! ways is possible and not 3!. Secondly You are only assuming C to sit in first place and D on right next to him. C can also take 2nd place, 3rd place and 4th place. So we have

C D _ _ (Here D is forced to right side hence 2 ways)
_ C D _ (Here can sit on right or left of C hence 2*2 = 4 ways)
_ _ C D (Here can sit on right or left of C hence 2*2 = 4 ways)
_ _ D C (Here D is forced to left side hence 2 ways)

Total 12 ways they can sit together. When deducted from 4! we get 24-12 = 12

Hope this clarifies

hi,
please check the ways here
C D X X
X C D X
X X C D
D C X X
X D C X
X X D C
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Re: Count number of ways to arrange 4 people A, B, C, D in a row so that [#permalink] New post   Updated on: 10 Aug 2011, 08:53
tranglenh wrote:
Sudhanshuacharya wrote:
tranglenh wrote:
Q: Count number of ways to arrange 4 people A, B, C, D in a row so that C, D not sit next to each other

[*] Manhattan solution: (Total number of arrangement - number C, D sit next)
Manhattan approach: pretend that C, D stuck together : then Count the number of ways 2 people not sitting next to each other,
Total number of arrangements: 4! = 24
then number of ways of arrangement so that C, D next to each other is: 3! = 6. Since C, D are distinct, so all number of ways C, D next to each other is: 6x2 = 12.
--> Number of arrangements C, D not sit next: 24 -12 = 12

[*]PFD: (Total number of arrangement - number C, D sit next)
PFD approach: Force C sit on D's right by creating 3 slots for D to sit in: seat 1,2,3. After D choose his seat, S automatically sit on his right.
So, total number of arrangement where C, D next to each other: 3x1 = 3 ( since D can sit on C's right --> then total arrangement is: 3x2 = 6

---> Number of arrangement C, D not sit next: 24 - 6 = 18 (page 83 PFD)
 :roll:
Please help to find the error here? why two answer different


Firstly when you are forcing D to sit on right of C, you are left with 2 seats hence only 2! ways is possible and not 3!. Secondly You are only assuming C to sit in first place and D on right next to him. C can also take 2nd place, 3rd place and 4th place. So we have

C D _ _ (Here D is forced to right side hence 2 ways)
_ C D _ (Here can sit on right or left of C hence 2*2 = 4 ways)
_ _ C D (Here can sit on right or left of C hence 2*2 = 4 ways)
_ _ D C (Here D is forced to left side hence 2 ways)

Total 12 ways they can sit together. When deducted from 4! we get 24-12 = 12

Hope this clarifies

hi,
please check the ways here
C D X X
X C D X
X X C D
D C X X
X D C X
X X D C


And: please notice that:

- First case: C to the right of D:
Select D first - 3 ways (since D can not sit in the last sit bcz C is always on his right)
Select C then - 1 way (the only next place to the right of D)
The number here is: 3x1 ( not 3!)
- Second case: C to the left of D
Same with first case --> 3x1
- Total arrangements if: 6 ways

Since we can check the number of ways by enumerating all cases above, I suppose there are problems with my understanding of probability in Manhattan approach ??? ( not the Probability for Dummies).

Please check!

Originally posted by tranglenh on 10 Aug 2011, 08:44.
Last edited by tranglenh on 10 Aug 2011, 08:53, edited 1 time in total.
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Re: Count number of ways to arrange 4 people A, B, C, D in a row so that [#permalink] New post   10 Aug 2011, 08:48
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tranglenh wrote:
Sudhanshuacharya wrote:
tranglenh wrote:
Q: Count number of ways to arrange 4 people A, B, C, D in a row so that C, D not sit next to each other

[*] Manhattan solution: (Total number of arrangement - number C, D sit next)
Manhattan approach: pretend that C, D stuck together : then Count the number of ways 2 people not sitting next to each other,
Total number of arrangements: 4! = 24
then number of ways of arrangement so that C, D next to each other is: 3! = 6. Since C, D are distinct, so all number of ways C, D next to each other is: 6x2 = 12.
--> Number of arrangements C, D not sit next: 24 -12 = 12

[*]PFD: (Total number of arrangement - number C, D sit next)
PFD approach: Force C sit on D's right by creating 3 slots for D to sit in: seat 1,2,3. After D choose his seat, S automatically sit on his right.
So, total number of arrangement where C, D next to each other: 3x1 = 3 ( since D can sit on C's right --> then total arrangement is: 3x2 = 6

---> Number of arrangement C, D not sit next: 24 - 6 = 18 (page 83 PFD)
 :roll:
Please help to find the error here? why two answer different


Firstly when you are forcing D to sit on right of C, you are left with 2 seats hence only 2! ways is possible and not 3!. Secondly You are only assuming C to sit in first place and D on right next to him. C can also take 2nd place, 3rd place and 4th place. So we have

C D _ _ (Here D is forced to right side hence 2 ways)
_ C D _ (Here can sit on right or left of C hence 2*2 = 4 ways)
_ _ C D (Here can sit on right or left of C hence 2*2 = 4 ways)
_ _ D C (Here D is forced to left side hence 2 ways)

Total 12 ways they can sit together. When deducted from 4! we get 24-12 = 12

Hope this clarifies

hi,
please check the ways here
C D X X
X C D X
X X C D
D C X X
X D C X
X X D C


Please distinguish each person with a different name. You'll see the problem. Only the pair that needs to be together may be considered a whole entity.

C D A B
C D B A
Are two different positions.

Moreover, when you consider CD as bundle say \(\theta\)

Now, \(\theta,A,B\) can be arranged in 3!=6 ways
And, the sub particles within \(\theta\) can be arranged in 2! ways.

Total=6*2=12(When sub-particles together)

When separated:

\(4!-12=12\)
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Re: Count number of ways to arrange 4 people A, B, C, D in a row so that [#permalink] New post   10 Aug 2011, 08:59
[/quote]
Please distinguish each person with a different name. You'll see the problem. Only the pair that needs to be together may be considered a whole entity.

C D A B
C D B A
Are two different positions.

Moreover, when you consider CD as bundle say \(\theta\)

Now, \(\theta,A,B\) can be arranged in 3!=6 ways
And, the sub particles within \(\theta\) can be arranged in 2! ways.

Total=6*2=12(When sub-particles together)

When separated:

\(4!-12=12\)[/quote]



Ah, yes, I got it now. Many Thanks for your help!

(But then the conclusion is that the solution in Probability for dummies is wrong!)
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Re: Count number of ways to arrange 4 people A, B, C, D in a row so that [#permalink] New post   10 Aug 2011, 10:55
a b cd

to arrange abcd its 4!

to arrange abcd so that cd will sit next to each other you treat cd as 1 unit. like a b cd so the number of arrangements of 3 items is 3!
now c and d can replace so it's 3!*2!

so the answer for not is:

4!-3!*2!

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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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Re: Count number of ways to arrange 4 people A, B, C, D in a row so that [#permalink]
10 Aug 2011, 10:55
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