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# There are 10 students named alphabetically from A to J.

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Math Expert
Joined: 02 Aug 2009
Posts: 5213

Kudos [?]: 5855 [1], given: 117

There are 10 students named alphabetically from A to J. [#permalink]

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31 Mar 2016, 05:01
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Difficulty:

55% (hard)

Question Stats:

42% (01:28) correct 58% (00:53) wrong based on 76 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) $$\frac{8}{9}$$
[Reveal] Spoiler: OA

_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 5855 [1], given: 117

SC Moderator
Joined: 13 Apr 2015
Posts: 1505

Kudos [?]: 1179 [0], given: 890

Location: India
Concentration: Strategy, General Management
WE: Information Technology (Consulting)
Re: There are 10 students named alphabetically from A to J. [#permalink]

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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

Kudos [?]: 1179 [0], given: 890

Math Expert
Joined: 02 Aug 2009
Posts: 5213

Kudos [?]: 5855 [0], given: 117

Re: There are 10 students named alphabetically from A to J. [#permalink]

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31 Mar 2016, 20:35
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

Hi Vyshak,..
Although approach is correct, Answer is wrong
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 5855 [0], given: 117

SC Moderator
Joined: 13 Apr 2015
Posts: 1505

Kudos [?]: 1179 [0], given: 890

Location: India
Concentration: Strategy, General Management
WE: Information Technology (Consulting)
Re: There are 10 students named alphabetically from A to J. [#permalink]

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31 Mar 2016, 21:29
chetan2u wrote:

Hi Vyshak,..
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.

Kudos [?]: 1179 [0], given: 890

Math Expert
Joined: 02 Aug 2009
Posts: 5213

Kudos [?]: 5855 [0], given: 117

Re: There are 10 students named alphabetically from A to J. [#permalink]

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31 Mar 2016, 21:48
Expert's post
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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
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 5855 [0], given: 117

Manager
Joined: 01 Mar 2014
Posts: 138

Kudos [?]: 10 [0], given: 616

Schools: Tepper '18
Re: There are 10 students named alphabetically from A to J. [#permalink]

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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.

Kudos [?]: 10 [0], given: 616

Math Expert
Joined: 02 Aug 2009
Posts: 5213

Kudos [?]: 5855 [0], given: 117

Re: There are 10 students named alphabetically from A to J. [#permalink]

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01 Apr 2016, 05:34
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
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 5855 [0], given: 117

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Re: There are 10 students named alphabetically from A to J. [#permalink]

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09 Oct 2017, 21:01
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Re: There are 10 students named alphabetically from A to J.   [#permalink] 09 Oct 2017, 21:01
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