## no two ladies sit together?

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### no two ladies sit together?

by sanju09 » Tue Apr 15, 2014 4:04 am
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

[spoiler]OA A[/spoiler]

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by [email protected] » Tue Apr 15, 2014 5:03 am
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])

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

Cheers,
Brent

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by Quasar Chunawala » Wed Apr 16, 2014 10:17 pm
The explanation above is extremely clear. It helps if you remember permutations and combinations from college-time.

The number of ways of arranging r objects from a set of n items are

The number of ways of selecting r objects from n choices are

You can arrange 5 gentlemen in P(5,5) = 5! = 120 ways

_M_M_M_M_M_

Further, you can arrange 4 women in the 6 empty slots in P(6,4) = 6!/2! = 360 ways

So, number of ways you can do this = (120)(360) = 43,200 ways

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by nikhilgmat31 » Thu Oct 08, 2015 1:35 am
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 .....

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by [email protected] » Thu Oct 08, 2015 6:43 am
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
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by nikhilgmat31 » Thu Oct 08, 2015 8:18 pm
[email protected] 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.

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by [email protected] » Thu Oct 08, 2015 9:30 pm
nikhilgmat31 wrote:
[email protected] 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.

In the scenario _M_M_M_M_M_ we can place the 4 women in ANY of the 6 spaces and we will meet the restriction that no 2 women can sit together.

Cheers,
Brent
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by sanju09 » Thu Oct 08, 2015 10:42 pm
Such a great explanation, Brent! Nothing is clearer than this, it's just just treat to watch. It's simply a practical and methodical approach to such problems, where formulaic approaches such as 9!/(5!*4!) = 126 ..... could make one confused whether it's a Permutation case or what? Formulaic approaches on most of the GMAT Math Problems on the real test such as this could take one to the wrong side of the road. Hence we should rely more on methodical, innovative, and practical approaches on hard problems in particular rather than looking for any magic stick. Majestic
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by ash4gmat » Tue May 03, 2016 6:24 am
[email protected] 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])

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

Cheers,
Brent

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)

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by [email protected] » Tue May 03, 2016 6:53 am
ash4gmat wrote:
[email protected] 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])

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

Cheers,
Brent