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Six children — A, B, C, D, E, and F — are going to sit in six chairs i

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Six children — A, B, C, D, E, and F — are going to sit in six chairs i [#permalink]

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New post 04 Mar 2015, 04:35
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A
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Six children — A, B, C, D, E, and F — are going to sit in six chairs in a row. Child E must be somewhere to the left of child F. How many possible configurations are there for the children?

A. 60
B. 180
C. 240
D. 360
E. 720



Kudos for a correct solution.

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Re: Six children — A, B, C, D, E, and F — are going to sit in six chairs i [#permalink]

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New post 04 Mar 2015, 04:39
Bunuel wrote:
Six children — A, B, C, D, E, and F — are going to sit in six chairs in a row. Child E must be somewhere to the left of child F. How many possible configurations are there for the children?

A. 60
B. 180
C. 240
D. 360
E. 720



Kudos for a correct solution.



Similar yet different questions:
seven-children-a-b-c-d-e-f-and-g-are-going-to-sit-in-seven-c-194099.html
seven-children-a-b-c-d-e-f-and-g-are-going-to-sit-in-seven-c-194097.html
seven-children-a-b-c-d-e-f-and-g-are-going-to-sit-in-seven-c-194096.html
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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: Six children — A, B, C, D, E, and F — are going to sit in six chairs i [#permalink]

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New post 04 Mar 2015, 11:50
1
Hi Bunuel,

I am getting the answer 220.

My answer is as follows,

F cannot be in first position:

If F is in second position; then E must be in First position- Remaining 4 children can be arranged in 4! ways= 24

If F is in Third position; then E can be two places, If we select 1 letter for left of F, then we will have 4C1X2X3=24

If F is in fourth position, then E can be in three places, So we have 4C2X3X2= 36

If F is in fifth position, then E can be in 4 places, 4c3X4X1= 16

If F is in sixth position, then we have 5! wasys= 120.

Summing all, I am getting 220 ways. Please highlight me where I am going wrong.

Bunuel wrote:
Six children — A, B, C, D, E, and F — are going to sit in six chairs in a row. Child E must be somewhere to the left of child F. How many possible configurations are there for the children?

A. 60
B. 180
C. 240
D. 360
E. 720



Kudos for a correct solution.

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Re: Six children — A, B, C, D, E, and F — are going to sit in six chairs i [#permalink]

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New post 04 Mar 2015, 12:18
1
Hi,

I am getting the answer as 360.

My answer is as follows,

F is in first position from the right, then the rest 5 positions can be filled in 5! ways=120

If F is in Second position; then the first position can be filled in 4 ways, and the rest 4 positions in 4! ways; i.e. 24*4=96

If F is in Third position; then the first and second position can be filled in 4 and 3 ways, and the rest 3 positions in 3! ways; i.e. 4*3*3!=72

If F is in Fourth position; then the first, second and the third position can be filled in 4, 3, and 2 ways ways, and the rest 2 positions in 2! ways; i.e. 4*3*2*2!=48

If F is in Fifth position; then the first, second third and the fourth position can be filled in 4, 3, 2 and 1 way, and the rest 1positions in 1! ways; i.e. 4!=24

Hence, answer should be 360.

Thanks
Bunuel wrote:
Six children — A, B, C, D, E, and F — are going to sit in six chairs in a row. Child E must be somewhere to the left of child F. How many possible configurations are there for the children?

A. 60
B. 180
C. 240
D. 360
E. 720



Kudos for a correct solution.
[/quote]
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Re: Six children — A, B, C, D, E, and F — are going to sit in six chairs i [#permalink]

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New post 04 Mar 2015, 12:42
12
3
The answer is D.... But, the way to reach to the correct answer is not as long as according to some guys !!!


Here is a simple rule ... First count the order of sitting people in row without any restriction....


As notice 6 people sitting in a ROW , so with no restriction the total number of arranging for 6 people is 6! = 720 ways.


So, in this step, lets apply the restriction ... restriction is that E must sit some where in the left of F...

Very simple rule : in these cases ONLY divide the total number of arrangement to 2 !!!! because in HALF of the cases E can sit in the left of F...

So the total number of arrangement is : 720/2 = 360 .... answer D


IMPORTANT notice : IF the problem said the 6 guys wanted to sit IN A CIRCULAR TABLE OR some thing like that, the total number of arrangement would be (6-1) = 5! = 120 and then 120/2

=60 :lol: :lol: :lol: :lol:
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Re: Six children — A, B, C, D, E, and F — are going to sit in six chairs i [#permalink]

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New post 04 Mar 2015, 12:46
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IMO D
Let us take a smaller example. Consider only A, B, C and assume that we need to find all possible arrangements when A is somewhere to the left of B.

All possible combinations: ABC, ACB, BAC, BCA, CAB, CBA

As you can see, exactly three of the cases are such when A is to the left of B. Meaning, half of the times, A will be to left of B. We can generalize this rule for A, B, C, D and we will find that in exactly 12 arrangements, A will be to the left of B.

Using the same concept here, we have six kids, so in all = \(6!\) i.e. \(720\) arrangements possible. Among these \(720\) combinations half of the times E will be somewhere to the left of F. Hence \(360\)
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Re: Six children — A, B, C, D, E, and F — are going to sit in six chairs i [#permalink]

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New post 04 Mar 2015, 19:11
1
Bunuel wrote:
Six children — A, B, C, D, E, and F — are going to sit in six chairs in a row. Child E must be somewhere to the left of child F. How many possible configurations are there for the children?

A. 60
B. 180
C. 240
D. 360
E. 720



Kudos for a correct solution.


A smart question that tests your presence of mind more than ability to count. Six students can be arranged in 6! ways, which gives us 720 arrangements. Of these, 360 arrangements will be such that E is to the left of F and the remaining 360 will have E to the right. Answer is D.
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Re: Six children — A, B, C, D, E, and F — are going to sit in six chairs i [#permalink]

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New post 04 Mar 2015, 21:52
2
1
Bunuel wrote:
Six children — A, B, C, D, E, and F — are going to sit in six chairs in a row. Child E must be somewhere to the left of child F. How many possible configurations are there for the children?

A. 60
B. 180
C. 240
D. 360
E. 720



Kudos for a correct solution.


Here is a post detailing the symmetry principle: http://www.veritasprep.com/blog/2011/10 ... s-part-ii/

Using the symmetry principle, you say that number of possible configurations = 6!/2 = 360

Answer (D)
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Re: Six children — A, B, C, D, E, and F — are going to sit in six chairs i [#permalink]

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New post 08 Mar 2015, 14:29
Bunuel wrote:
Six children — A, B, C, D, E, and F — are going to sit in six chairs in a row. Child E must be somewhere to the left of child F. How many possible configurations are there for the children?

A. 60
B. 180
C. 240
D. 360
E. 720



Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

If we wanted, we could make this one extremely difficult, counting out all kinds of possibilities in several different cases. Instead, we are going to make this ridiculously easy.

First of all, with absolutely no restrictions, how many ways can the six children be arranged on the six chairs? That’s a permutation of the 6 items —- 6P6 = 6! = 720. That’s the total number of arrangements with no restrictions. Of course, those 720 arrangements have all kinds of symmetry to them. In particular, in all of those arrangements overall, it’s just as likely for E to be to the left of F as it is for E to be to the right of F. Therefore, exactly half must have E to the right of F, and exactly half must have E to the left of F. Therefore, exactly (1/2)*720 = 360 of the arrangements have E to the left of F.

Answer = D.
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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

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Six children — A, B, C, D, E, and F — are going to sit in six chairs i [#permalink]

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New post 07 Mar 2018, 20:20
1
Hi All,

This question can be solved in a couple of different ways. Here's a more drawn-out explanation for the correct answer:

We're given 6 spots:

_ _ _ _ _ _

And we have to put a child in each spot; E MUST be to the left of F though. Here's one possible way that this can occur:

E F _ _ _ _

In this option, any of the remaining 4 children can be placed in each of the 4 remaining spots, so we have….

E F (4)(3)(2)(1) = 24 options with E and F in the 1st and 2nd spots, respectively.

This pattern will occur over-and-over. For example….If E and F were in other places….

(4)E(3)(2)F(1) = 24 options with E and F in the 2nd and 5th spots, respectively.

So we really just need all of the possible placements for E and F, then we can multiply that result by 24…

While this is not necessarily the most efficient way to approach this task, the work isn't that hard….

E F _ _ _ _
E _ F _ _ _
E _ _ F _ _
E _ _ _ F _
E _ _ _ _ F

_ E F _ _ _
_ E _ F _ _
_ E _ _ F _
_ E _ _ _ F

_ _ E F _ _
_ _ E _ F _
_ _ E _ _ F

_ _ _ E F _
_ _ _ E _ F

_ _ _ _ E F

15 possible placements for E and F (given the restriction that E must be to the left of F).

15 x 24 = 360

Final Answer:

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Re: Six children — A, B, C, D, E, and F — are going to sit in six chairs i   [#permalink] 07 Mar 2018, 20:20
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