no two ladies sit together?

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

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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Source: Beat The GMAT — Problem Solving | Tue Apr 15, 2014 4:04 am

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » 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])

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 Hanneson - Creator of GMATPrepNow.com

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Quasar Chunawala: Newbie | Next Rank: 10 Posts; Posts: 7; Joined: Wed Apr 16, 2014 9:41 pm; Thanked: 4 times

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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nikhilgmat31: Legendary Member; Posts: 518; Joined: Tue May 12, 2015 8:25 pm; Thanked: 10 times

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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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » 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

Brent Hanneson - Creator of GMATPrepNow.com

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nikhilgmat31: Legendary Member; Posts: 518; Joined: Tue May 12, 2015 8:25 pm; Thanked: 10 times

by nikhilgmat31 » Thu Oct 08, 2015 8:18 pm

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.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Thu Oct 08, 2015 9:30 pm

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.

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

Brent Hanneson - Creator of GMATPrepNow.com

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

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

The mind is everything. What you think you become. -Lord Buddha

Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

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ash4gmat: Senior | Next Rank: 100 Posts; Posts: 57; Joined: Thu Sep 17, 2015 2:57 am

by ash4gmat » Tue May 03, 2016 6:24 am

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)

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Tue May 03, 2016 6:53 am

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)

By the time we complete stage 5, we have seated all of the people. So, we have completed our task of seating everyone.

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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