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# A group of four women and three men have tickets for seven

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Joined: 10 Jun 2007
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A group of four women and three men have tickets for seven [#permalink]  15 Aug 2007, 18:17
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!
Director
Joined: 11 Jun 2007
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Re: Theater Seats [#permalink]  15 Aug 2007, 19:45
bkk145 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!

going to try to work it out even though I am probably wrong
different possibilities - we know that the men and women will alternate seats since men will not sit next to each other
1) M W M W M W W

2) W M W M W M W

3) W W M W M W M

I think the answer is E
Director
Joined: 03 May 2007
Posts: 888
Schools: University of Chicago, Wharton School
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Re: Theater Seats [#permalink]  15 Aug 2007, 21:02
bkk145 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!

C.
= 7! - 6x5x4!
= 7! - 6!
= 7! - 5!3x2
= 7! - 5!3!
Manager
Joined: 08 Oct 2006
Posts: 215
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Re: Theater Seats [#permalink]  15 Aug 2007, 22:00
Fistail wrote:
bkk145 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!

C.
= 7! - 6x5x4!
= 7! - 6!
= 7! - 5!3x2
= 7! - 5!3!

can you explain this bit.

my way is
total ways =7!
ways that men can sit together with women=5! ways. considering men as a single seat.
3!= no. of ways men can sit together.

So 7!- 5!.3!.
Director
Joined: 03 May 2007
Posts: 888
Schools: University of Chicago, Wharton School
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Kudos [?]: 75 [0], given: 7

Re: Theater Seats [#permalink]  16 Aug 2007, 06:37
excelgmat wrote:
Fistail wrote:
bkk145 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!

C.
= 7! - 6x5x4!
= 7! - 6!
= 7! - 5!3x2
= 7! - 5!3!

can you explain this bit.

my way is
total ways =7!
ways that men can sit together with women=5! ways. considering men as a single seat.
3!= no. of ways men can sit together.

So 7!- 5!.3!.

required = total - 3 men adjecent

total = 7!
3 men adjacent to each other can be placed in 5 places = 5
3 men can be placed adjacent to each other in a place = 3! = 3x2 = 6 ways
remaining 4 people can be placed = 4! ways

so, required = 7! - 6x5x4! = 7! - 6! = 7! - 5!3!
Re: Theater Seats   [#permalink] 16 Aug 2007, 06:37
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