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PS-Permutations (m08q30) [#permalink]
 02 Dec 2008, 07:23
Question Stats:
69% (01:39) correct
30% (01:13) wrong based on 65 sessions
If among 5 children there are 2 siblings, in how many ways can the children be seated in a row so that the siblings do not sit together? (A) 38 (B) 46 (C) 72 (D) 86 (E) 102 Source: GMAT Club Tests - hardest GMAT questions
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Re: PS-Permutations [#permalink]
02 Dec 2008, 08:54
C
!5 - Total ways
2!4 - If sibblings sit together
!5-2!4 = 72
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Re: PS-Permutations [#permalink]
02 Dec 2008, 09:28
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Agree with HG.
Total ways = 5! =120
The two siblings can sit among themselves in 2 ways and taken together as one, then 4 people can sit in 4! = 24 ways.
Hence total = 120 - (24 *2) = 120 - 48 = 72 ways.
Whats the QA ?
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Re: PS-Permutations (m08q30) [#permalink]
27 Aug 2010, 09:08
What if the problem was how many arrangements are possible if 2 sibling were not to sit together on a circular bench. Solution :- sit together : (4-1)!x2!=3.2.2 12 ways total circular arrangements : (5-1)!=4! 24ways Not sitting together : 24-12 = 12 ways Is this right ......
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Re: PS-Permutations (m08q30) [#permalink]
27 Aug 2010, 16:33
All,
I may be getting my basics wrong coz . Can you check my thought process out
5 people can sit in 5!=120 ways
now if we consider 5 seats with s1 and s2 as siblings and x as other people we have following arrangements
s1 s2 x x x
x s1 s2 x x
x x s1 s2 x
x x x s1 s2
A total of 4 and since s1 and s2 can interchange we have possible ways to sit as 2*4
so total is 120-8
obviously this is wrong but i cant fathom the reason
any help is appreciated
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Re: PS-Permutations (m08q30) [#permalink]
27 Aug 2010, 17:33
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someonear wrote: All,
I may be getting my basics wrong coz . Can you check my thought process out
5 people can sit in 5!=120 ways
now if we consider 5 seats with s1 and s2 as siblings and x as other people we have following arrangements
s1 s2 x x x
x s1 s2 x x
x x s1 s2 x
x x x s1 s2
A total of 4 and since s1 and s2 can interchange we have possible ways to sit as 2*4
so total is 120-8
obviously this is wrong but i cant fathom the reason
any help is appreciated Ofcorse its wrong and even you know it. and the reason for this is bacause here you are only considering the sitting arrangements of s1 and s2, but what about those three x's which are all different. those 3 different x's can be arranged in 3! ways which is 6 and then u multiply by 8 which is 48. 120-48 = 72. Ideally I would use this method: No restrictions: 5!ways = 120 With restrictions- when two of them are always together: 4!*2! = 48 Therefore, required answer = 120- 48 = 72
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Re: PS-Permutations (m08q30) [#permalink]
30 Aug 2010, 10:02
vivektripathi wrote: If among 5 children there are 2 siblings, in how many ways can the children be seated in a row so that the siblings do not sit together? (A) 38 (B) 46 (C) 72 (D) 86 (E) 102 Source: GMAT Club Tests - hardest GMAT questions C
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Re: PS-Permutations (m08q30) [#permalink]
30 Aug 2010, 21:30
how many questions we get in Permutations, combinations in real GMAT?
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Re: PS-Permutations (m08q30) [#permalink]
31 Aug 2010, 14:45
srivicool wrote: how many questions we get in Permutations, combinations in real GMAT? you should expect 1-2 questions from each topic..... 1-2 max probability, 1-2 MAX combinations, etc Unless you have absolutely MASTERED topics such as algebra, arithmetic (ie factors, LCM), inequalities etc which are absolutely basic, my advice would be to stick to the basics for combinametrics...... more than happy to elaborate more if you'd like
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Re: PS-Permutations (m08q30) [#permalink]
01 Sep 2010, 03:09
thanks for the explanation zisis!!!
please elaborate if you have more info..
thanks
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Re: PS-Permutations (m08q30) [#permalink]
03 Oct 2010, 01:56
if i say the 48(the number of combination that two siblings will sit together) is from
first sit is 2 (choose 1 sibling out of 2)
second sit is 1 (there is only one sibling left to choose)
third sit is 3 (there are 3 other people who are not sibling can choose from)
forth sit is 2 (there are 3 other people who are not sibling can choose from)
fifth sit is 1 (.....)
(2*1*3*2*1)*4
the 4 is from:
s1 s2 x x x
x s1 s2 x x
x x s1 s2 x
x x x s1 s2
is this a better way to explain?
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Re: PS-Permutations (m08q30) [#permalink]
18 Jan 2011, 07:35
shekharvineet wrote: someonear wrote:
Ideally I would use this method:
No restrictions: 5!ways = 120
With restrictions- when two of them are always together: 4!*2! = 48
Therefore, required answer = 120- 48 = 72 This seams to be an good method, unfortunately I don´t seem to understand why I put 4!*2!, but not 5!*2! one of these days  thanks
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Re: PS-Permutations [#permalink]
18 Jan 2011, 07:39
gameCode wrote: The two siblings can sit among themselves in 2 ways and taken together as one, then 4 people can sit in 4! = 24 ways.
I Got it ! I love this forum !
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Re: PS-Permutations (m08q30) [#permalink]
01 Sep 2011, 10:23
easy one...C is the answer
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Re: PS-Permutations (m08q30) [#permalink]
01 Sep 2011, 12:10
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Re: PS-Permutations (m08q30) [#permalink]
03 Sep 2011, 08:46
It's C. Although it took me sometime getting it. But it definitely isn't among the hardest like you rightly put it.
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Re: PS-Permutations [#permalink]
06 Sep 2011, 13:52
gameCode wrote: Agree with HG.
Total ways = 5! =120
The two siblings can sit among themselves in 2 ways and taken together as one, then 4 people can sit in 4! = 24 ways.
Hence total = 120 - (24 *2) = 120 - 48 = 72 ways.
Whats the QA ? This is the way I solve it.
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Re: PS-Permutations (m08q30) [#permalink]
14 Sep 2011, 02:47
Guys, what if there are 3 siblings? the combinations of sitting together would be 3! (3!) or 3! (3) ??
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Re: PS-Permutations (m08q30) [#permalink]
29 Aug 2012, 23:13
krishnasty wrote: Guys, what if there are 3 siblings?
the combinations of sitting together would be 3! (3!) or 3! (3) ?? 3! (3!) is correct. Approach should be- Treat 3 Siblings as one, hence now total no. of kids would be 3. So 3 kids will be seated in 3! way, while 3 siblings can arrange themselves in 3! ways. So total ways would be 3!.3!
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Re: PS-Permutations (m08q30) [#permalink]
03 Sep 2012, 07:18
If among 5 children there are 2 siblings, in how many ways can the children be seated in a row so that the siblings do not sit together? (A) 38 (B) 46 (C) 72 (D) 86 (E) 102 Total no.of ways in which the children could be seated so that the siblings do not sit together equals to ( = )Total no.of seating for 5 children minus (-) Total number of seating possible when the children could be seated so that the siblings do sit together =5! - (2!4!) =120-48 =72
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Re: PS-Permutations (m08q30)
[#permalink]
03 Sep 2012, 07:18
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