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There are 10 students named alphabetically from A to J.

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chetan2u
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There are 10 students named alphabetically from A to J. [#permalink] There are 10 students named alphabetically from A to J.   31 Mar 2016, 05:01
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A
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C
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E

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55% (01:38) correct 45% (01:38) wrong based on 127 sessions
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There are 10 students named alphabetically from A to J. What is the Probability that A and D do not sit together if all 10 sit around a circular table?

A) \(\frac{2}{9}\)

B) \(\frac{2}{5}\)

C) \(\frac{7}{9}\)

D) \(\frac{4}{5}\)

E) \(\frac{8}{9}\)
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C
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Re: There are 10 students named alphabetically from A to J. [#permalink] There are 10 students named alphabetically from A to J.   31 Mar 2016, 11:09
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Editing my solution:

Number of students = 10

Number of ways 10 students can sit around a circular table = (10 - 1)! = 9!

Number of ways A and D sit together (consider A and D as one entity) = (9 - 1)! = 8! * 2

Number of ways A and D do not sit together = 9! - (8! * 2)

Probability = (9! - (8! * 2))/9! = 1 - 2/9 = 7/9

Answer: C
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Re: There are 10 students named alphabetically from A to J. [#permalink] There are 10 students named alphabetically from A to J.   31 Mar 2016, 20:35
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Vyshak wrote:
Number of students = 10

Number of ways 10 students can sit around a circular table = (10 - 1)! = 9!

Number of ways A and D sit together (consider A and D as one entity) = (9 - 1)! = 8!

Number of ways A and D do not sit together = 9! - 8!

Probability = (9! - 8!)/9! = 1 - 1/9 = 8/9

Answer: E


Hi Vyshak,..
Relook into your solution closely...
Although approach is correct, Answer is wrong
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Re: There are 10 students named alphabetically from A to J. [#permalink] There are 10 students named alphabetically from A to J.   31 Mar 2016, 21:29
chetan2u wrote:

Hi Vyshak,..
Relook into your solution closely...
Although approach is correct, Answer is wrong


Thanks Chetan. I have corrected my solution. I had a confusion whether to take the inner arrangement of A and D when I first solved the question. I was under the impression that AD and DA are identical arrangements. I, now, realized that its not.
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Re: There are 10 students named alphabetically from A to J. [#permalink] There are 10 students named alphabetically from A to J.   31 Mar 2016, 21:48
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chetan2u wrote:
There are 10 students named alphabetically from A to J. What is the Probability that A and D do not sit together if all 10 sit around a circular table?

A) \(\frac{2}{9}\)

B) \(\frac{2}{5}\)

C) \(\frac{7}{9}\)

D) \(\frac{4}{5}\)

E) \(\frac{8}{9}\)


.


Hi,

Since it is circular Permutation, ways= (n-1)!
so the 10 students can be seated in 10-1=9!..

We have to find ways A and D do not seat together..
It is easier to find ways both will seat together..
let A and D be one person. so total person =9..
these 9 can be seated in 9-1=8!

BUT A and D can be seated in 2! within themselves..
so TOTAL ways = 2*8!..

So PROBABILITY that A and D sit together = \(2*\frac{8!}{9!}= \frac{2}{9}\)

Therefore Prob that both do not sit together = \(1-\frac{2}{9}=\frac{7}{9}\)

C
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Re: There are 10 students named alphabetically from A to J. [#permalink] There are 10 students named alphabetically from A to J.   01 Apr 2016, 04:44
chetan2u wrote:
chetan2u wrote:
There are 10 students named alphabetically from A to J. What is the Probability that A and D do not sit together if all 10 sit around a circular table?

A) \(\frac{2}{9}\)

B) \(\frac{2}{5}\)

C) \(\frac{7}{9}\)

D) \(\frac{4}{5}\)

E) \(\frac{8}{9}\)


.


Hi,

Since it is circular Permutation, ways= (n-1)!
so the 10 students can be seated in 10-1=9!..

We have to find ways A and D do not seat together..
It is easier to find ways both will seat together..
let A and D be one person. so total person =9..
these 9 can be seated in 9-1=8!

BUT A and D can be seated in 2! within themselves..
so TOTAL ways = 2*8!..

So PROBABILITY that A and D sit together = \(2*\frac{8!}{9!}= \frac{2}{9}\)

Therefore Prob that both do not sit together = \(1-\frac{2}{9}=\frac{7}{9}\)

C


I did not do this using the formula for circular permutations and took the total no. of ways to be 10! Does this formula hold true in all circular arrangements. I would really appreciate if you could explain as to how we can derive this. Apologies if this is a very basic question but just wanted to make sure i have got the concepts right. Thanks for your help.
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Re: There are 10 students named alphabetically from A to J. [#permalink] There are 10 students named alphabetically from A to J.   01 Apr 2016, 05:34
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MeghaP wrote:
chetan2u wrote:
chetan2u wrote:
There are 10 students named alphabetically from A to J. What is the Probability that A and D do not sit together if all 10 sit around a circular table?

A) \(\frac{2}{9}\)

B) \(\frac{2}{5}\)

C) \(\frac{7}{9}\)

D) \(\frac{4}{5}\)

E) \(\frac{8}{9}\)


.


Hi,

Since it is circular Permutation, ways= (n-1)!
so the 10 students can be seated in 10-1=9!..

We have to find ways A and D do not seat together..
It is easier to find ways both will seat together..
let A and D be one person. so total person =9..
these 9 can be seated in 9-1=8!

BUT A and D can be seated in 2! within themselves..
so TOTAL ways = 2*8!..

So PROBABILITY that A and D sit together = \(2*\frac{8!}{9!}= \frac{2}{9}\)

Therefore Prob that both do not sit together = \(1-\frac{2}{9}=\frac{7}{9}\)

C


I did not do this using the formula for circular permutations and took the total no. of ways to be 10! Does this formula hold true in all circular arrangements. I would really appreciate if you could explain as to how we can derive this. Apologies if this is a very basic question but just wanted to make sure i have got the concepts right. Thanks for your help.


Hi,
the formula for circular permutations is (n-1)! as against n! in a row AND it is valid for all circular arrangements AND it also finds its place in ACTUAL GMAT.

NOW WHY of it?
If you are seating 4 people in a row, you can seat them in 4! ways..
However in circular permutations it becomes (4-1)=3!...
When you shift each person in a ROW by one place, it is a different arrangement
for example a,b,c,d is different from d,a,b,c or c,d,a,b or b,c,d,a..

But it is not the case in circular permutations/arrangements..
When you shift them by one seat each, the relative position of each other does NOT change..
Draw a circle place a,b,c,d AND then d,a,b,c-- you will find they are the same
this is the reason for (n-1)! as against n! in a row
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Re: There are 10 students named alphabetically from A to J. [#permalink] There are 10 students named alphabetically from A to J.   18 Feb 2024, 05:36
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