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help for circular permutation

Expert replies
by sana.noor » Sun Aug 11, 2013 1:46 am
There are 6 people at a party sitting at a round table with 6 seats: A,B,C,D,E AND F. A CANNOT sit next to either D or F. How many ways can the 6 people be seated?

A. 720
B. 120
C. 108
D. 84
E. 48

OA is C
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Source: — Problem Solving |

by GMATGuruNY » Sun Aug 11, 2013 3:22 am
sana.noor wrote:There are 6 people at a party sitting at a round table with 6 seats: A,B,C,D,E AND F. A CANNOT sit next to either D or F. How many ways can the 6 people be seated?

A. 720
B. 120
C. 108
D. 84
E. 48

OA is C
For circular arrangements, we count the number of ways to arrange the REMAINING people RELATIVE to the first person seated.

Once A is seated:
Number of options for the seat to A's left = 3. (Of the 5 remaining people, anyone but D or F.)
Number of options for the seat to A's right = 2. (Of the 4 remaining people, anyone but D or F.)
Number of ways to arrange the remaining 3 people = 3! = 6.
To combine these options, we multiply:
3*2*6 = 36.

The correct answer does not appear among the answer choices.
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by hemant_rajput » Sun Aug 11, 2013 8:28 am
Another approach -

First fix A.

Now total possible arrangement - 5! = 120

Arrangements when D and F sit next to = make D and F sit next to A * the rest of available seats with rest of the members = 2! * 3! = 12

120 - 12 = 108

Answer c
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by GMATGuruNY » Sun Aug 11, 2013 12:42 pm
hemant_rajput wrote:Another approach -

First fix A.

Now total possible arrangement - 5! = 120

Arrangements when D and F sit next to = make D and F sit next to A * the rest of available seats with rest of the members = 2! * 3! = 12

120 - 12 = 108

Answer c
You have correctly subtracted from the total Bad Case 1: the arrangements in which BOTH D AND F sit next to A.
But we must also subtract from the total Bad Case 2: the arrangements in which EITHER D OR F -- but NOT both -- sit next to A.

Bad Case 2: Either D or F (but not both) next to A
Number of options for the disallowed person sitting next to A = 2. (Either D or F.)
Number of seat options for the disallowed person = 2. (To the right or left of A.)
Number of options for the OTHER seat next to A = 3. (Anyone but D or F, since only one of them can sit next to A.)
Number of ways to arrange the remaining 3 people = 3!.
To combine these options, we multiply:
2*2*3*3! = 72.

Thus:
Total favorable arrangements = total possible arrangements - bad case 1 - bad case 2 = 120-12-72 = 36.
Last edited by GMATGuruNY on Mon Aug 12, 2013 7:37 am, edited 1 time in total.
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I unlock the best way for YOU to solve problems.

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by hemant_rajput » Mon Aug 12, 2013 5:40 am
Oh shit, I totally miss that part.

Thanks a million Mitch.
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