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vivektripathi · **Posted:** Tue Dec 02, 2008 6:23 am

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

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gameCode · **Posted:** Tue Dec 02, 2008 8:28 am

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 ?

shekharvineet · **Posted:** Fri Aug 27, 2010 4:33 pm

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

HG · **Posted:** Tue Dec 02, 2008 7:54 am

C

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

!5-2!4 = 72

vivektripathi · **Posted:** Wed Dec 03, 2008 5:06 am

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 ?

OA is C

srivicool · **Posted:** Fri Aug 27, 2010 7:05 am

Thanks for the explanation. can this problem be solved in any other way?

gsothee · **Posted:** Fri Aug 27, 2010 8:08 am

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 ......

shekharvineet · **Posted:** Fri Aug 27, 2010 9:35 am

gsothee wrote:

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 ......

Yah thats correct

someonear · **Posted:** Fri Aug 27, 2010 3:33 pm

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

amankalra · **Posted:** Sat Aug 28, 2010 12:01 am

C.

srivicool · **Posted:** Sat Aug 28, 2010 3:45 am

thanks for the explanation shekarvineet

zisis · **Posted:** Mon Aug 30, 2010 9:02 am

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

srivicool · **Posted:** Mon Aug 30, 2010 8:30 pm

how many questions we get in Permutations, combinations in real GMAT?

zisis · **Posted:** Tue Aug 31, 2010 1:45 pm

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

FQ · **Posted:** Tue Aug 31, 2010 1:50 pm

C --- 5! - 4! * 2

srivicool · **Posted:** Wed Sep 01, 2010 2:09 am

thanks for the explanation zisis!!!

please elaborate if you have more info..

thanks

alan7858 · **Posted:** Sun Oct 03, 2010 12:56 am

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?

hafgola · **Posted:** Tue Jan 18, 2011 6:35 am

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

hafgola · **Posted:** Tue Jan 18, 2011 6:39 am

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 !

rongali · **Posted:** Thu Sep 01, 2011 9:23 am

easy one...C is the answer

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