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by armaan700+ » Tue Jan 26, 2010 10:02 am
A group of four women and three men have tickets for seven adjacent seats in one row of a theatre. If the three men will not sit in three adjacent seats, how many possible different seating arrangements are there for these 7 theatre-goers?

(A) 7! - 2!3!2!
(B) 7! - 4!3!
(C) 7! - 5!3!
(D) 7 × 2!3!2!
(E) 2!3!2!
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by ajith » Tue Jan 26, 2010 11:25 am
armaan700+ wrote:A group of four women and three men have tickets for seven adjacent seats in one row of a theatre. If the three men will not sit in three adjacent seats, how many possible different seating arrangements are there for these 7 theatre-goers?

(A) 7! - 2!3!2!
(B) 7! - 4!3!
(C) 7! - 5!3!
(D) 7 × 2!3!2!
(E) 2!3!2!
We will find the total number of arrangements and deduct from it the combination 3 men sitting together

3 mean sitting together can be considered as a block and we have 5 blocks which can be arranged in 5! ways further these 3 mean can be arranged among themselves in 3! ways making the ways in which 3 mean sit together 3!5!


So the num of arrangements in which 3 men do not sit together = 7!-5!3!
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by kewltot » Tue Jan 26, 2010 12:01 pm
A group of four women and three men have tickets for seven adjacent seats in one row of a theatre. If the three men will not sit in three adjacent seats, how many possible different seating arrangements are there for these 7 theatre-goers?


total number of ways 7 seats are occupied are 7! ways

Number of ways all 3 men can seat together = 5! ways

Number of ways all 3 men can interchange their seats = 3!

Total number of ways three men will not sit in three adjacent seats = 7! - 5! * 3!
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by hai1 » Wed Jan 27, 2010 2:59 am
"3 mean sitting together can be considered as a block and we have 5 blocks which can be arranged in 5! "

Why can 3 men sit together only in 5! ways? How is this deduced?
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by ajith » Wed Jan 27, 2010 4:26 am
hai1 wrote:"3 mean sitting together can be considered as a block and we have 5 blocks which can be arranged in 5! "

Why can 3 men sit together only in 5! ways? How is this deduced?
Initially we have 7 spots and 7 men who can be arranged in 7! ways


Now, if we are forcibly making three of them stay together in one order, and there are 4 other people

we can have = 5! ways to arrange them (4 men and a block of three men)


Say we Have people A B C D E F and G

Initial arrangement doesnt force anybody to stay together s

But in the latter arrangement we are asking say, CDE to stay together and CDE is considered one block and there are a total of 5! arrangement.

In turn CDE can be arranged among themselves in 3! ways (CDE, CED, DCE, DEC, ECD, EDC) (for the next part)

Total no of arrangements in which 3 persons (CDE) stay together is 5!*3!
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