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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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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



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


Check out the blog posts on the link given in my signature below.

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

Answer (D)

Originally posted by KarishmaB on 04 Mar 2015, 21:52.
Last edited by KarishmaB on 17 Oct 2022, 01:50, edited 1 time in total.
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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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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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Re: Six children — A, B, C, D, E, and F — are going to sit in six chairs i [#permalink]
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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]
Expert Reply
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


If there is no restriction, the number of seating arrangements is 6! = 720. Of these 720 arrangements, half of them will have E sitting to the left of F (and the other half will have E sitting to the right of F). Therefore, the number of possible seating arrangements with E is to the left of F is 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]
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.
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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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