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Count number of ways to arrange 4 people A, B, C, D in a row so that 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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Count number of ways to arrange 4 people A, B, C, D in a row so that 10 Aug 2011, 08:16
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tranglenh
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
Show more

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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Count number of ways to arrange 4 people A, B, C, D in a row so that 10 Aug 2011, 08:35
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Sudhanshuacharya
tranglenh
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
Show more
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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Count number of ways to arrange 4 people A, B, C, D in a row so that Updated on: 10 Aug 2011, 08:53
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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tranglenh
Sudhanshuacharya
tranglenh
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
Show more

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!
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Count number of ways to arrange 4 people A, B, C, D in a row so that 10 Aug 2011, 08:48
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tranglenh
Sudhanshuacharya
tranglenh
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
Show more

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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Count number of ways to arrange 4 people A, B, C, D in a row so that 10 Aug 2011, 08:59
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[/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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Count number of ways to arrange 4 people A, B, C, D in a row so that 10 Aug 2011, 10:55
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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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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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