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

 It is currently 16 Jul 2018, 05:56

### 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

# Nine family members: 5 grandchildren (3 brothers and 2 siste

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 47015
Nine family members: 5 grandchildren (3 brothers and 2 siste [#permalink]

### Show Tags

25 Jul 2014, 09:09
1
15
00:00

Difficulty:

95% (hard)

Question Stats:

54% (02:07) correct 46% (02:00) wrong based on 316 sessions

### HideShow timer Statistics

Nine family members: 5 grandchildren (3 brothers and 2 sisters) and their 4 grandparents are to be seated around a circular table. How many different seating arrangements are possible so that 2 sisters are seated between any two of the three brothers?

A. 120
B. 480
C. 1440
D. 2880
E. 8640

Kudos for a correct solution.

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 47015
Nine family members: 5 grandchildren (3 brothers and 2 siste [#permalink]

### Show Tags

25 Jul 2014, 09:10
8
4
SOLUTION

Nine family members: 5 grandchildren (3 brothers and 2 sisters) and their 4 grandparents are to be seated around a circular table. How many different seating arrangements are possible so that 2 sisters are seated between any two of the three brothers?

A. 120
B. 480
C. 1440
D. 2880
E. 8640

Consider two brothers and two sisters between them as one unit: {BSSB}.

So, now we have 6 units: {G}, {G}, {G}, {G}, {B}, and {BSSB}.

These 6 units can be arranged around a circular table in (6-1)!=5! ways.

Next, analyze {BSSB} unit:
We can choose 2 brothers out of 3 for the unit in $$C^2_3=3$$ ways;
These brothers, within the unit, can be arranged in 2! ways: $$\{B_1, S, S, B_2\}$$ or $$\{B_2, S, S, B_1\}$$
The sisters, within the unit, also can be arranged in 2! ways: $$\{B, S_1, S_2, B\}$$ or $$\{B, S_2, S_1, B\}$$

Therefore, the final answer is 5!*3*2*2=1440.

Try NEW Combinations DS question.
_________________
Manager
Joined: 04 Sep 2012
Posts: 99
Location: Philippines
Concentration: Marketing, Entrepreneurship
Schools: Ross (Michigan) - Class of 2017
GMAT 1: 620 Q48 V27
GMAT 2: 660 Q47 V34
GMAT 3: 700 Q47 V38
GPA: 3.25
WE: Sales (Manufacturing)
Re: Nine family members: 5 grandchildren (3 brothers and 2 siste [#permalink]

### Show Tags

25 Jul 2014, 18:17
4
1
Bunuel wrote:

Nine family members: 5 grandchildren (3 brothers and 2 sisters) and their 4 grandparents are to be seated around a circular table. How many different seating arrangements are possible so that 2 sisters are seated between any two of the three brothers?

A. 120
B. 480
C. 1440
D. 2880
E. 8640

Kudos for a correct solution.

Number of ways to arrange the 2 sisters = $$2$$
Number of ways to choose the 2 brothers out of 3 = $$3C2 = 3$$
Number of ways to arrange the 2 sisters between the 2 brothers (only the brothers switch places since the sisters have to be in the middle) = $$2$$
Number of ways to arrange the 2 sisters between the 2 brothers with the leftover brother and 4 grand parents around a table = $$(6-1)! = 5!$$

$$2x3x2x5! = 12*120=1440$$
##### General Discussion
Intern
Joined: 22 Jul 2011
Posts: 24
Location: Bulgaria
Re: Nine family members: 5 grandchildren (3 brothers and 2 siste [#permalink]

### Show Tags

26 Jul 2014, 05:56
What am I doing wrong? I got 480...

So, around the table, we have: {brother, sister, sister, brother}, and then 5 more family members in random order (because the 3rd brother doesn't have to sit next to his siblings, right?)

So, we have 4 possible arrangements of the first 4 siblings: 2x2x1x1=4

And the five other family members (brother + 4 grandparents) can sit in 5!=120 possible arrangements.

Overall, all possible arrangements are therefore (2)(2)(5!)=(4)(120)=480

So, why not B?
Math Expert
Joined: 02 Sep 2009
Posts: 47015
Re: Nine family members: 5 grandchildren (3 brothers and 2 siste [#permalink]

### Show Tags

29 Jul 2014, 14:01
2
SOLUTION

Nine family members: 5 grandchildren (3 brothers and 2 sisters) and their 4 grandparents are to be seated around a circular table. How many different seating arrangements are possible so that 2 sisters are seated between any two of the three brothers?

A. 120
B. 480
C. 1440
D. 2880
E. 8640

Consider two brothers and two sisters between them as one unit: {BSSB}.

So, now we have 6 units: {G}, {G}, {G}, {G}, {B}, and {BSSB}.

These 6 units can be arranged around a circular table in (6-1)!=5! ways.

Next, analyze {BSSB} unit:
We can choose 2 brothers out of 3 for the unit in $$C^2_3=3$$ ways;
These brothers, within the unit, can be arranged in 2! ways: $$\{B_1, S, S, B_2\}$$ or $$\{B_2, S, S, B_1\}$$
The sisters, within the unit, also can be arranged in 2! ways: $$\{B, S_1, S_2, B\}$$ or $$\{B, S_2, S_1, B\}$$

Therefore, the final answer is 5!*3*2*2=1440.

Kudos points given to correct solutions.

Try NEW Combinations DS question.
_________________
Manager
Joined: 23 Jan 2012
Posts: 71
Re: Nine family members: 5 grandchildren (3 brothers and 2 siste [#permalink]

### Show Tags

30 Sep 2014, 16:55
Bunuel why did you do (6-1)!? We have 6 units, so couldn't they be arranged in 6! ways?
Manager
Status: I am not a product of my circumstances. I am a product of my decisions
Joined: 20 Jan 2013
Posts: 127
Location: India
Concentration: Operations, General Management
GPA: 3.92
WE: Operations (Energy and Utilities)
Re: Nine family members: 5 grandchildren (3 brothers and 2 siste [#permalink]

### Show Tags

30 Sep 2014, 21:30
3
p2bhokie wrote:
Bunuel why did you do (6-1)!? We have 6 units, so couldn't they be arranged in 6! ways?

This is because it is a case of "CIRCULAR PERMUTATION"

Suppose if 5 Boys, A, B,C, D and E are standing in a straight line and we have an arrangement ABCDE. Now, if we move each boy one place to the right then arrangement will become EABCD. ABCDE and EABCD are not the same. So we calculate N! or 5! arrangements when dealing with straight line arrangements.

In the above case, we are dealing with Circular arrangements. Now suppose the same 5 Boys are seated around a circular table, then we have an arrangement ABCDE. If we move each boy one place to the right the arrangement will still be ABCDE. So we are counting the same arrangement more than once and this would be wrong. So we subtract the duplicates from N! and hence we get (N-1) ! arrangements.

Hope this helps !!!
Manager
Joined: 23 Jan 2012
Posts: 71
Re: Nine family members: 5 grandchildren (3 brothers and 2 siste [#permalink]

### Show Tags

01 Oct 2014, 05:37
Thanks Bunuel. The explanation definitely helps. Thanks again.

Posted from my mobile device
Current Student
Joined: 25 Nov 2014
Posts: 102
Concentration: Entrepreneurship, Technology
GMAT 1: 680 Q47 V38
GPA: 4
Re: Nine family members: 5 grandchildren (3 brothers and 2 siste [#permalink]

### Show Tags

30 Mar 2015, 01:58
2 sisters => 2!.
sisters between any 2 brothers => 1) arrangement of 2 brothers out of 3 where order matters => 3P2 = 3!.
leaving aside 2 S and 2B(chosen one's), we have 1B and 4GPs and one combined unit of BssB => 6 units to be arranged.
Circular => 5!.
ans = pdt of all = 2! * 3! * 5! = 120 * 12 = 1440.
Ans C.
_________________

Kudos!!

Retired Moderator
Joined: 29 Apr 2015
Posts: 863
Location: Switzerland
Concentration: Economics, Finance
Schools: LBS MIF '19
WE: Asset Management (Investment Banking)
Re: Nine family members: 5 grandchildren (3 brothers and 2 siste [#permalink]

### Show Tags

28 Sep 2015, 13:13
2
I solved it by drawing a little illustration:

Attachment:

Unbenannt.jpg [ 25.31 KiB | Viewed 3561 times ]

There is one restriction, so that the two sister should be seated between two brothers. So first ignore the two sisters and calculate the circular permutation of n = 7 (9-2 sisters).

circular permutation of 7 = (7-1)! = 6! = 720

Now the two sisters can sit in 2! ways

Hence, total combinations: 2*720 = 1440
_________________

Saving was yesterday, heat up the gmatclub.forum's sentiment by spending KUDOS!

PS Please send me PM if I do not respond to your question within 24 hours.

Manager
Joined: 13 Apr 2015
Posts: 75
Concentration: General Management, Strategy
GMAT 1: 620 Q47 V28
GPA: 3.25
WE: Project Management (Energy and Utilities)
Re: Nine family members: 5 grandchildren (3 brothers and 2 siste [#permalink]

### Show Tags

04 Oct 2015, 00:31
Can any one tell what is wron in my aproach.

Total ways of seating 9 people in a circle is 8!

And ways in which the sisters will not be inbetween brothers is when all three brothers are together. I.e. (6-1) 5!.

Now subtracting, 8! - 5! should't give the ways in which 2 sisters are seated bw brothers..?
Retired Moderator
Joined: 29 Apr 2015
Posts: 863
Location: Switzerland
Concentration: Economics, Finance
Schools: LBS MIF '19
WE: Asset Management (Investment Banking)
Re: Nine family members: 5 grandchildren (3 brothers and 2 siste [#permalink]

### Show Tags

04 Oct 2015, 00:52
goldfinchmonster wrote:
Can any one tell what is wron in my aproach.

Total ways of seating 9 people in a circle is 8!

And ways in which the sisters will not be inbetween brothers is when all three brothers are together. I.e. (6-1) 5!.

Now subtracting, 8! - 5! should't give the ways in which 2 sisters are seated bw brothers..?

With your approach you are dealing with big numbers which is imo not very efficient. Mind that the wys in which the sisters do not sit together is not complete when all brothers sit together. You can have e.g. the following combination: BRO SIS BRO BRO SIS GP GP GP GP --> Now does the Brothers sit together? No! Are the two sisters in between them? No!

Try to visualise and for this example look at the graph I did above, it may help you?
_________________

Saving was yesterday, heat up the gmatclub.forum's sentiment by spending KUDOS!

PS Please send me PM if I do not respond to your question within 24 hours.

Non-Human User
Joined: 09 Sep 2013
Posts: 7250
Re: Nine family members: 5 grandchildren (3 brothers and 2 siste [#permalink]

### Show Tags

04 Mar 2018, 09:37
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: Nine family members: 5 grandchildren (3 brothers and 2 siste   [#permalink] 04 Mar 2018, 09:37
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®.