nikhilgmat31 wrote:Hi Brent,

Counting the number of way for 5 Men = 120 seems fine.

But Why can't we have a pattern of 4 women as M_M_M_M_M

where we will have 4 * 3 * 2 * 1 = 24 options for ladies.

total of 120* 24 = 2880 options.

Also can't we solve this question using other formula as - 9!/(5!*4!) = 126 .....

Also can't we solve this question using other formula as - 9!/(5!*4!) = 126 .....

nikhilgmat31 wrote:

Brent@GMATPrepNow wrote:

nikhilgmat31 wrote:Hi Brent,

Counting the number of way for 5 Men = 120 seems fine.

But Why can't we have a pattern of 4 women as M_M_M_M_M

where we will have 4 * 3 * 2 * 1 = 24 options for ladies.

total of 120* 24 = 2880 options.

Also can't we solve this question using other formula as - 9!/(5!*4!) = 126 .....

The pattern you describe (M_M_M_M_M) is already included in my solution.
We can use your pattern but we must recognize that we still need to examine other patterns (like MM_M_M_M_ and _M_MM_M_M) if want to count ALL possibilities.

My solution considers all possibilities.

Also can't we solve this question using other formula as - 9!/(5!*4!) = 126 .....

This formula finds the number of ways to select 5 seats from 9 seats. It does not adhere to the restriction that no two women can sit together.

Cheers,
Brent

Yes Brent,
I still didn't get how your solution all the patterns like _M_M_M_M_M_ or M_M_M_M_M or _MM_M_M_M or MM_M_M_M_ or _M_MM_M_M.

Please help me understand.

ash4gmat wrote:

Brent@GMATPrepNow wrote:

sanju09 wrote:In how many ways can 4 ladies and 5 gentlemen be seated in a row so that no two ladies sit together?
A. 43200
B. 21600
C. 5760
D. 2880
E. 1440

Take the task of seating everyone and break it into stages.

Stage 1: Arrange all of the men in a row
We can arrange k unique objects in k! ways.
Since there are 5 men, we can arrange them in 5! (120 ways)

IMPORTANT: Now place an empty chair on either side of each man as follows:
_M_M_M_M_M_

Note: This prevents the women from sitting together because there is now a man separating each of 6 empty chairs.

Stage 2: Seat a woman
There are 6 seats, so we can complete this stage in 6 ways

Stage 3: Seat another woman
There are 5 seats remaining, so we can complete this stage in 5 ways

Stage 4: Seat another woman
There are 4 seats remaining, so we can complete this stage in 4 ways

Stage 5: Seat the last woman
There are 3 seats remaining, so we can complete this stage in 3 ways

By the Fundamental Counting Principle (FCP), we can complete all 5 stages (and thus seat all 9 people) in (120)(6)(5)(4)(3) ways ([spoiler]= 43200, ways[/spoiler])

Answer: A

Here's a similar question to practice with: https://www.beatthegmat.com/p-c-pls-help-t29328.html

Cheers,
Brent

Aside: For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775

Brent why we have stopped with (120)(6)(5)(4)(3) and not included another 2 ways .

What I mean is (120)(6)(5)(4)(3)(2)