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M08-30

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M08-30 [#permalink]

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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

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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

Without limitations, 5 children can be seated in \(5! = 120\) ways. Find the number of ways to seat the 5 children so that the siblings DO sit together. 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 \(4!*2 = 48\). Thus, the number of combinations in which the siblings DO NOT sit together is 120 - 48 = 72.

Answer: C
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M08-30 [#permalink]

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New post 30 Apr 2016, 16:14
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
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Re: M08-30 [#permalink]

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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..
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Re: M08-30 [#permalink]

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New post 15 May 2016, 18:55
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= 120-48= 72

I hope it helps,
best regards.
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M08-30 [#permalink]

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New post 28 Sep 2016, 21:11
I applied different approach -



there are 5 seats , 2 siblings S1 and S2 and three others A,B,C

________ ________________ _____________ __________ ______________
---5 ways ----------3 ways----------- 3 ways--------2 ways------1 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
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Re: M08-30 [#permalink]

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New post 20 Oct 2016, 18:01
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
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Re: M08-30 [#permalink]

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New post 10 Dec 2017, 04:46
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
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Re: M08-30 [#permalink]

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New post 10 Dec 2017, 04:54
abhinashgc wrote:
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


How can you consider siblings to be identical and indistinguishable?

Yes, the number of ways to arrange AABCD is 5!/2! but how does this give the arrangements when A's are together? The arrangements will be:
AABCD
ABACD
ABCAD
ABCDA
...
BCADA
BCDAA
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M08-30   [#permalink] 10 Dec 2017, 04:54
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