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
OA
D
Six children — A, B, C, D, E, and F
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- gmat_guy666
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First, let's IGNORE the rule about child E seated somewhere to the left of child F.gmat_guy666 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
When we ignore the rule, we can arrange the 6 children in 6! ways (= 720 ways)
Now recognize that, in HALF of those 720 arrangements, child E is seated somewhere to the LEFT of child F, and in the other HALF of those 720 arrangements, child E is seated somewhere to the RIGHT of child F.
So, the number of arrangements where child E is seated somewhere to the left of child F = (1/2)720 = 360
Answer: D
NOTE: This question is almost identical to this one: https://www.beatthegmat.com/counting-six ... tml#198825
Cheers,
Brent
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Here are a few more to practice with:
- https://www.beatthegmat.com/permutation-t261691.html
- https://www.beatthegmat.com/vowels-and-c ... 49946.html
- https://www.beatthegmat.com/how-many-pos ... 77574.html
Cheers,
Brent
- https://www.beatthegmat.com/permutation-t261691.html
- https://www.beatthegmat.com/vowels-and-c ... 49946.html
- https://www.beatthegmat.com/how-many-pos ... 77574.html
Cheers,
Brent
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Hi gmat_guy666,
Brent's explanation is arguably the most efficient way to get to the correct answer. 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
Brent's explanation is arguably the most efficient way to get to the correct answer. 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
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One more approach:gmat_guy666 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
There are no constraints on A, B, C, and D.
Number of options for A = 6. (Any of the 6 chairs.)
Number of options for B = 5. (Any of the 5 remaining chairs.)
Number of options for C = 4. (Any of the 4 remaining chairs.)
Number of options for D = 3. (Any of the 3 remaining chairs.)
Of the 2 remaining chairs, the one most to the left must be occupied by E, since E must sit to the left of F.
Number of options for E = 1. (The chair most to the left.)
Number of options for F = 1. (Only 1 chair remains.)
To combine the options above, we multiply:
6*5*4*3*1*1 = 360.
The correct answer is D.
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An Alternate methods is:
Select two places for B and E and there is only 1 way for them to sit on those chair, the chairs for B and E can be selected in 6C2 ways = 15
Now remaining 4 chairs can be occupied by other 4 people in 4! ways = 24 ways
Total ways : 6C2 x 4! = 15 x 24 = 360
Answer: Option D
Select two places for B and E and there is only 1 way for them to sit on those chair, the chairs for B and E can be selected in 6C2 ways = 15
Now remaining 4 chairs can be occupied by other 4 people in 4! ways = 24 ways
Total ways : 6C2 x 4! = 15 x 24 = 360
Answer: Option D
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