combination,table, 700+, manhattan

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combination,table, 700+, manhattan

by soudeh » Thu Oct 26, 2017 12:57 pm
In how many ways can 6 people be seated at a round table if one of those seated cannot sit next to 2 of the other 5?

a) 720
b) 120
c) 108
d) 84
e) 48

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by GMATGuruNY » Thu Oct 26, 2017 1:52 pm
I believe that the following reflects the intent of the problem:
In how many ways can 6 people A, B, C, D, E and F be seated at a round table if A cannot sit next to B or C?

a) 720
b) 120
c) 108
d) 84
e) 48
To count circular arrangements:
1. Place someone in the circle.
2. Count the number of ways to arrange the REMAINING people.

After A is placed in the circle:
Number of options for the seat to the left of A = 3. (Of the 5 remaining people, anyone but B or C.)
Number of options for the seat to the right of A = 2. (Of the 4 remaining people, anyone but B or C.)
Number of ways to arrange the remaining 3 people = 3! = 6.
To combine these options, we multiply:
3*2*6 = 36.

The correct number of arrangements is not among the answer choices.
What is the source of this problem?
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Re: combination,table, 700+,

by GMATGuruNY » Thu Oct 26, 2017 2:09 pm
Given the answer choices, I suspect that the prompt is intended to ask the following:
In how many ways can 6 people A, B, C, D, E and F be seated at a round table if A, B and C cannot sit in 3 adjacent seats such that A is between B and C?

a) 720
b) 120
c) 108
d) 84
e) 48
As noted in my post directly above, to count circular arrangements:
1. Place someone in the circle.
2. Count the number of ways to arrange the remaining people.

Good arrangements = (total possible arrangements) - (bad arrangements).

Total possible arrangements:
After A has been seated at the table, the number of ways to arrange the remaining 5 people = 5! = 120.

Bad arrangements:
In a bad arrangement, A, B and C sit in 3 adjacent seats, with A between B and C.
After A has been seated:
Number of options for the seat to the left of A = 2. (B or C.)
Number of options for the seat to the right of A = 1. (Must be B or C, whoever is not seated to the left of A.)
Number of ways to arrange the remaining 3 people = 3! = 6.
To combine these options, we multiply:
2*1*6 = 12.

Thus:
Good arrangements = 120 - 12 = 108.

The correct answer is C.

The problem as written seems poorly worded.
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I unlock the best way for YOU to solve problems.

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