abhi332 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!

[spoiler]
OA:C[/spoiler]

Let us make these 3 men sit together it can be done in 3! ways
Now let us arrange them into the gaps between 4 women - There are 5 gaps MMMGGGG, GMMMGGG, GGMMMMGG, GGGMMMG, GGGGMMM) they can be arranged in 5! ways

Total no of ways to arrange 7 people = 7!

Total no of ways in which 3 men sit adjacent to each other = 3!*5!

No of ways in which they dont sit adjacent to each other = 7! -3!*5!

abhi332 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!

[spoiler]
OA:C[/spoiler]

Total = 7!
Now,let the four women be arranged in 4!ways.
So, men in 5C3 x 3! ways
So, answer will be 60 x 4!

Alternatively,
taking out the unwanted cases,
let the three men be considered a separate entity.
So,total 4women + 1 entity - Arrangement in 5! ways.
Men can interchange b/w themselves in 3! ways.
[spoiler]Hence 7! - 5!*3! which is C[/spoiler]