November 14, 2018 November 14, 2018 07:00 PM PST 08:00 PM PST Join the webinar and learn timemanagement tactics that will guarantee you answer all questions, in all sections, on time. Save your spot today! Nov. 14th at 7 PM PST November 15, 2018 November 15, 2018 10:00 PM MST 11:00 PM MST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth $100 with the 3 Month Pack ($299)
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 50572

Question Stats:
77% (01:02) correct 23% (02:09) wrong based on 102 sessions
HideShow timer Statistics



Math Expert
Joined: 02 Sep 2009
Posts: 50572

Re M0830
[#permalink]
Show Tags
15 Sep 2014, 23:38



Senior Manager
Joined: 18 Aug 2014
Posts: 324

Bunuel wrote: The siblings can be regarded as one unit so there are 4! combinations. But within this unit the siblings can sit in two different ways. So the number of ways to seat the 5 children so that the siblings DO sit together is \(4!*2 = 48\).
I thought 4! included all the possibilities not just a particular order? So that there is no need to multiply the 4! by 2. Like how many ways can you arrange a, b, and c 3! = 6, as below shows if B and C were siblings both placements are accounted for are they not? A B C A C B B A C B C A C A B C B A
_________________
Please help me find my lost Kudo's bird



Math Expert
Joined: 02 Aug 2009
Posts: 7025

Re: M0830
[#permalink]
Show Tags
30 Apr 2016, 20:11
redfield wrote: Bunuel wrote: The siblings can be regarded as one unit so there are 4! combinations. But within this unit the siblings can sit in two different ways. So the number of ways to seat the 5 children so that the siblings DO sit together is \(4!*2 = 48\).
I thought 4! included all the possibilities not just a particular order? So that there is no need to multiply the 4! by 2. Like how many ways can you arrange a, b, and c 3! = 6, as below shows if B and C were siblings both placements are accounted for are they not? A B C A C B B A C B C A C A B C B A Hi, the 2 is for the two siblings who have been taken as 1..
The siblings can be regarded as one unit so there are 4! combinations.. Let these siblings be \(S_1\) and \(S_2\) AND others be A,B and C.. so one of the arrangements out of 4! is \(A,B,C,S_1,S_2\)... But we can arrange the siblings in two ways here.. second would be \(A,B,C,S_2,S_1\)...Similarily there will be two ways for each ARRANGEMENT.. that is why 4!*2..
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
GMAT online Tutor



Current Student
Joined: 12 Nov 2015
Posts: 59
Location: Uruguay
Concentration: General Management
GMAT 1: 610 Q41 V32 GMAT 2: 620 Q45 V31 GMAT 3: 640 Q46 V32
GPA: 3.97

Re: M0830
[#permalink]
Show Tags
15 May 2016, 17:55
On the other hand if we don't see the cunning plan of treating the siblings as one unit, we could do as follows:
Sibling 1 and Sibling 2 occupy 2 spaces out of 5, there are 4 ways for them to sit together, in each of these 4 ways they can switch position> S1,S2,A,B,C; S2,S1,A,B,C>> the 3 children that are not siblings are arranged as follows: 3! (since the kids occupy the 3 lest over spaces). the 2 siblings can be arranged in 2 ways giving us a total of 3*2*2=12 ways to sit the children for the first arrangement.. that includes S1,S2,B,C,D.. S2,S1,A,C,B (for example)... If you do the same calculus for all the positions where the siblings sit close (so.. A,S1,S2,C,B... A,B,S1,S2,C.. A,B,C,S1,S2) the result is 48.
Total arrangements possible : 5!  48= 12048= 72
I hope it helps, best regards.



Intern
Joined: 11 Nov 2014
Posts: 35
Concentration: Marketing, Finance
WE: Programming (Computer Software)

I applied different approach  there are 5 seats , 2 siblings S1 and S2 and three others A,B,C ________ ________________ _____________ __________ ______________ 5 ways 3 ways 3 ways2 ways1 way  (not counting one sibling) 5*3*3*2*1 = 90 order of seating can be in any way. can you please tell me where am I wrong ? Bunuel



Intern
Joined: 01 Sep 2016
Posts: 12

Re: M0830
[#permalink]
Show Tags
20 Oct 2016, 17:01
Simpler way to do this: We have 3 children and two siblings
Arrange 3 children in \(3!\) ways
Consider the remaining two children: they have 4 possible places to be placed (ie, anywhere in _, where X represents the other 3 children): _ X _ X _ X _
Hence we can arrange the two siblings in 4P2 ways = \(\frac{4!}{(4  2)!}\)
Hence total probability is \(3! * \frac{4!}{(4  2)!}\) = \((3*2* 1) * \frac{(4*3*2*1)}{(2*1)}\) = \(6 * 12\) = \(72\)
This can be generalized for N children & X siblings



Intern
Joined: 15 Aug 2013
Posts: 40

Re: M0830
[#permalink]
Show Tags
10 Dec 2017, 03:46
Hi Bunuel , I got your explanation. But i tried with below approach,and answer was wrong.Could you please clarify where i was wrong No of way to arrange 5 where 2 are one kind 5!/2! Total no of arrangement for 5 =5! So Answer should be 5!5!/2!=60 Thanks



Math Expert
Joined: 02 Sep 2009
Posts: 50572

Re: M0830
[#permalink]
Show Tags
10 Dec 2017, 03:54



Manager
Joined: 11 Sep 2013
Posts: 157
Concentration: Finance, Finance

Re: M0830
[#permalink]
Show Tags
24 Jul 2018, 02:09
ABCSS
Let S sits first for the 2nd chair we have 3 choices. For the 3rd chair we have 3 and for the 4th chair we have 2 choices = 1*3*3*2*1 = 18
Now we can Put first S in any 4 chairs. So we have 4*18 = 72 choices










