In how many ways can Ann, Bea, Cam, Don, Ella and Fey be seated if Ann and Bea cannot be seated next to each other?
(A) 240
(B) 360
(C) 480
(D) 600
(E) 720
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Ann, Bea, Cam, Don, Ella and Fey be seated
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- sanju09
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- Rahul@gurome
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Total ways of seating 6 people = 6!
Consider Ann and Bea as one person, then number of ways of arranging 5 people = 5!
Number of ways of arranging Ann and Ben = 2!
Required number of ways of seating 6 people if Ann and Bea are not seated next to each other = 6! - 5!2! = 720 - 240 = 480
The correct answer is [spoiler](C)[/spoiler].
Consider Ann and Bea as one person, then number of ways of arranging 5 people = 5!
Number of ways of arranging Ann and Ben = 2!
Required number of ways of seating 6 people if Ann and Bea are not seated next to each other = 6! - 5!2! = 720 - 240 = 480
The correct answer is [spoiler](C)[/spoiler].
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Answer is 480
Solution: All 6 can occupy 6 places in 6! ways.
Now lets assume, Ann and Bea always sit together and consider them as one unit. So 5 people can occupy 5 places in 5! ways. Also Ann and Bea (whom we have considered one unit) can occupy the two places next to each other in 2 ways. Hence the total no. of ways when Ann and Bea sit together = 2*5!.
So, the no. of ways where Ann and Bea do not sit together =(total no of ways)- (no. of ways when they always sit together) = 6!-2*5! = 480
Solution: All 6 can occupy 6 places in 6! ways.
Now lets assume, Ann and Bea always sit together and consider them as one unit. So 5 people can occupy 5 places in 5! ways. Also Ann and Bea (whom we have considered one unit) can occupy the two places next to each other in 2 ways. Hence the total no. of ways when Ann and Bea sit together = 2*5!.
So, the no. of ways where Ann and Bea do not sit together =(total no of ways)- (no. of ways when they always sit together) = 6!-2*5! = 480
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Here's the long form approach that I finally figured out. Takes too long on the test, but it helped me visualize the problem.
We have 6! ways of arranging the people, so that's our starting point -> 720
I've setup a chart like so - If we have A and B assigned, we have 4 x 3 x 2 x 1 options with the remaining seats. So each row has 24 arrangements that we'll need to exclude.
1 2 3 4 5 6
A B 4 3 2 1
B A 4 3 2 1
4 A B 3 2 1
4 B A 3 2 1
4 3 A B 2 1
4 3 B A 2 1
4 3 2 A B 1
4 3 2 B A 1
4 3 2 1 A B
4 3 2 1 B A
Since there are ten total rows * 24 exclusions per row, we get 240 exclusions.
We have 6! ways of arranging the people, so that's our starting point -> 720
I've setup a chart like so - If we have A and B assigned, we have 4 x 3 x 2 x 1 options with the remaining seats. So each row has 24 arrangements that we'll need to exclude.
1 2 3 4 5 6
A B 4 3 2 1
B A 4 3 2 1
4 A B 3 2 1
4 B A 3 2 1
4 3 A B 2 1
4 3 B A 2 1
4 3 2 A B 1
4 3 2 B A 1
4 3 2 1 A B
4 3 2 1 B A
Since there are ten total rows * 24 exclusions per row, we get 240 exclusions.
- sanju09
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Why can't we stick with the shortest method as suggested by Rahul? After all this much knowledge of Permutations is expected of all GMAT aspirants. On GMAT, a logic is less qualified if it's longer than an already existing logic known to all.
Thanks
Thanks
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com