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# PS-Permutations (m08q30)

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Senior Manager
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02 Dec 2008, 07:23
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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

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions
Manager
Joined: 02 Nov 2008
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02 Dec 2008, 08:54
C

!5 - Total ways
2!4 - If sibblings sit together

!5-2!4 = 72
Manager
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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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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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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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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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Joined: 16 Feb 2010
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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

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

$$5! - 2(4!) = 120 - 2(24) = 120 - 48 = 72$$

C
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30 Aug 2010, 21:30
how many questions we get in Permutations, combinations in real GMAT?
Manager
Joined: 16 Feb 2010
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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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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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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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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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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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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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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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02 Sep 2013, 05:30
Another way to solve the problem - there are 3 children another than the siblings. If they sit in any order there will be a 4 places where we can adjust the siblings and thereby they won't be sitting together.
Total ways = (4C2)(2)(3!)

where,
4C2 - selections of any 2 - arrangement of siblings. spots out of 4
2 - arrangement of siblings
3! - arrangement of the 3 children.

Posted from my mobile device
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03 Sep 2013, 01:50
total number of ways = 5! = 120
2 siblings cannot sit together.So consider them as 1 unit.
total ways in which 4 people can sit = 4! = 24
the 2 siblings can sit among themselves in 2! ways.

5!-4!2! = 72

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29 May 2014, 12:13
I don't know why I am odd one out since I didn't get the solution as stated in previous posts.

Consider S1,S2 to be siblings and S3,S4 another.

Then people saying !4. But what if I put them as shown below:

S4..S2..S3..S1 (And surely this is one of the arrangement which we will get through !4).

So, isn't it wrong to count it.

Rgds,
TGC!
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29 May 2014, 12:26
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TGC wrote:
I don't know why I am odd one out since I didn't get the solution as stated in previous posts.

Consider S1,S2 to be siblings and S3,S4 another.

Then people saying !4. But what if I put them as shown below:

S4..S2..S3..S1 (And surely this is one of the arrangement which we will get through !4).

So, isn't it wrong to count it.

Rgds,
TGC!

Don't know which solution you are referring to but notice that there are 5 children, not 4.

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

{# of arrangement where the siblings do not sit together} = {total # of arrangements of 5 children} - {# of arrangements where the siblings sit together}.

{total # of arrangements of 5 children} = 5! = 120.

{# of arrangements where the siblings sit together}:
Consider two siblings as one unit {S1, S2}. In this case 4 units {S1, S2}, {X}, {Y}, {Z} can be arrangement in 4! ways. Siblings within their unit can be arranged in 2 ways: {S1, S2} or {S2, S1}. Hence # of arrangements where the siblings sit together is 4!*2 = 48.

{# of arrangement where the siblings do not sit together} = 120 - 48 = 72.

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30 May 2014, 00:45
Oh mistook 2 siblings for 2 pair of siblings.

Thanks for the example shown.

Rgds,
TGC!
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Re: PS-Permutations (m08q30)   [#permalink] 30 May 2014, 00:45

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