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

Question

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

mehrdadtaheri92 · Accepted Answer

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:

rockstar23 · Answer

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

Bunuel · Answer

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

shriramvelamuri · Answer

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.

aayush91 · Answer

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

aviram · Answer

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.

KarishmaB · Answer

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)

Bunuel · Answer

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.

EMPOWERgmatRichC · Answer

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:  D

GMAT assassins aren't born, they're made, 
Rich

ScottTargetTestPrep · Answer

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

gmatpro15 · Answer

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.

RishiRamdani · Answer

In how many ways can I place 2 persons on 6 chairs while keeping the order person 1 somewhere to the left of person 2 ?
6C2 = 6*5/2 = 15

In how many ways can I place the 4 persons remaining ?
4*3*2*1 = 24

Total # of configurations:
15 * 24 = 360
Answer D

Can someone confirm the reasoning above is correct ?

pierjoejoe · Answer

here we have to consider all the combinations where E is on the left of F. how to do that?
we can count the number of ways E and F can be seated in 6 seats. we want the order to NOT matter, because we want these 2 arrangements (for example):
EF _ _ _ _ (OK)
FE _ _ _ _ (NOT OK)
to be counted ONCE --> in this way we can immagine that we are only counting the once in which the condition is satisfied (E is on the left of F).

so we have 6C2 = 15 arrangements of E and F that differ at least by ONE different element (or disposition):
E F _ _ _ _ 
E _ F _ _ _ 
_ E F _ _ _ 
_ E _ F _ _ 
ecc..

for each one of these arrangements we can arrange the other 4 friends in 4 seats in 4! ways

total = 6C2 * 4! = 15 * 24 = 360

bumpbot · Answer

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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

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