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Theater arrangements

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Theater arrangements

by beat_gmat_09 » Tue Nov 09, 2010 8:53 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!

Pls help.
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by Dani@MasterGMAT » Tue Nov 09, 2010 9:11 am
use "total number of arrangements" - "forbidden options": those options where the three men DO sit together. Calculate each separately, then subtract.

Total number of arrangements = 7! ways to seat 7 people, regardless of limitations.

forbidden options: number of ways where the men DO sit together. Use a two step method for solving these "proximity" problems:
1) lump the men together into a single "entity" and figure out the number of ways of arranging the 4 women+single entity of "3 men": total of 5 "entities", or 5! different arrangements.

2) Think about the internal order of the men entity. There are 5!=120 different ways of placing arranging 4 women+entity, but the 3 men have 3! ways of arranging between them. For each of the 120 arrangements, you will get 3*2*1=6 arrangements depending on the internal arrangements of men within the entity. Thus, the final number of ways of arranging the 7 people with the men together is 5!*3!

Now, get back to our original calculation: "total number of arrangements" - "forbidden options = 7!-5!3! - or answer choice C.
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