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Permutation: girls not sitting together

by sevenplus » Wed Jul 07, 2010 10:43 am
In how many ways can 6 boys and 2 girls be seated in 8 chairs if the two girls can not be seated next to each other.
A) 40313
B) 40312
C) 35280
D) 5040
E) 720

The way I approached this problem:
Total number of ways in which both boys and girls can be seated = 8! = 40320
Total number of ways in which two girls can sit next to each other = 7!2! = 10080
Total number of ways in which two girls can NOT sit next to each other = 40320 - 10080 = 30240

Can someone help me find out what I am doing wrong here as I don't see 30240 as one of the answer choices?
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by kvcpk » Wed Jul 07, 2010 11:01 am
I believe what you have done is right.. What is the OA and source?
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by sevenplus » Wed Jul 07, 2010 11:06 am
Source is one of the test prep companies and OA is C. I also think that OA is wrong.
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by arora007 » Wed Jul 07, 2010 11:09 am
C is actually ( 8! - 7! ) , they might have forgotten about the 2! ways in which the two girls can sit besides eachother...
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by GMATGuruNY » Wed Jul 07, 2010 2:42 pm
sevenplus wrote:In how many ways can 6 boys and 2 girls be seated in 8 chairs if the two girls can not be seated next to each other.
A) 40313
B) 40312
C) 35280
D) 5040
E) 720

The way I approached this problem:
Total number of ways in which both boys and girls can be seated = 8! = 40320
Total number of ways in which two girls can sit next to each other = 7!2! = 10080
Total number of ways in which two girls can NOT sit next to each other = 40320 - 10080 = 30240

Can someone help me find out what I am doing wrong here as I don't see 30240 as one of the answer choices?
Your approach is sound. To cut down on the arithmetic, you should hold off multiplying until the final calculation:

8! - (7! * 2) = 7! (8 - 2) = 7! * 6 = 5040 * 6 = 30240.
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