GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 16 Jul 2018, 16:43

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Aug 2009
Posts: 6210
There are 10 students named alphabetically from A to J. [#permalink]

### Show Tags

31 Mar 2016, 05:01
1
3
00:00

Difficulty:

55% (hard)

Question Stats:

44% (01:24) correct 56% (00:55) wrong based on 80 sessions

### HideShow timer Statistics

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) $$\frac{8}{9}$$

_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

GMAT online Tutor

SC Moderator
Joined: 13 Apr 2015
Posts: 1712
Location: India
Concentration: Strategy, General Management
GMAT 1: 200 Q1 V1
GPA: 4
WE: Analyst (Retail)
Re: There are 10 students named alphabetically from A to J. [#permalink]

### Show Tags

31 Mar 2016, 11:09
1
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

Math Expert
Joined: 02 Aug 2009
Posts: 6210
Re: There are 10 students named alphabetically from A to J. [#permalink]

### Show Tags

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
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

GMAT online Tutor

SC Moderator
Joined: 13 Apr 2015
Posts: 1712
Location: India
Concentration: Strategy, General Management
GMAT 1: 200 Q1 V1
GPA: 4
WE: Analyst (Retail)
Re: There are 10 students named alphabetically from A to J. [#permalink]

### Show Tags

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.
Math Expert
Joined: 02 Aug 2009
Posts: 6210
Re: There are 10 students named alphabetically from A to J. [#permalink]

### Show Tags

31 Mar 2016, 21:48
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
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

GMAT online Tutor

Manager
Joined: 01 Mar 2014
Posts: 129
Schools: Tepper '18
Re: There are 10 students named alphabetically from A to J. [#permalink]

### Show Tags

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.
Math Expert
Joined: 02 Aug 2009
Posts: 6210
Re: There are 10 students named alphabetically from A to J. [#permalink]

### Show Tags

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
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

GMAT online Tutor

Non-Human User
Joined: 09 Sep 2013
Posts: 7262
Re: There are 10 students named alphabetically from A to J. [#permalink]

### Show Tags

09 Oct 2017, 21:01
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: There are 10 students named alphabetically from A to J.   [#permalink] 09 Oct 2017, 21:01
Display posts from previous: Sort by

# Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.