Last visit was: 24 Apr 2026, 16:05 It is currently 24 Apr 2026, 16:05
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 24 Apr 2026
Posts: 109,818
Own Kudos:
Given Kudos: 105,873
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,818
Kudos: 811,077
 [56]
4
Kudos
Add Kudos
52
Bookmarks
Bookmark this Post
avatar
shariq41
Joined: 23 Apr 2017
Last visit: 30 Jan 2018
Posts: 2
Own Kudos:
6
 [4]
Given Kudos: 177
Posts: 2
Kudos: 6
 [4]
1
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
User avatar
jokschmer
Joined: 20 Sep 2015
Last visit: 18 Sep 2018
Posts: 42
Own Kudos:
Given Kudos: 41
Status:Profile 1
GMAT 1: 690 Q48 V37
GPA: 3.2
WE:Information Technology (Finance: Investment Banking)
GMAT 1: 690 Q48 V37
Posts: 42
Kudos: 43
Kudos
Add Kudos
Bookmarks
Bookmark this Post
avatar
manishtank1988
Joined: 14 Oct 2012
Last visit: 31 Oct 2019
Posts: 112
Own Kudos:
287
 [2]
Given Kudos: 1,023
Products:
Posts: 112
Kudos: 287
 [2]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
shariq41
Bunuel
A few men and a few women are seated in a row. The number of men is one greater than the number of women. What is the total number of people?

(1) The people are seated so that no two women sit beside one another.
(2) The number of ways of arranging the seating positions of the men and women is 3(5!)(7!).


Answer = C

Explanation :-
Let the number of women be x. Therefore, the number of men is x+1.

1) We cannot find the total number of people just by using Statement 1.
For example :-
i) There are 3 men and 2 women such that : MWMMW.
ii) There are 4 men and 3 women such that : MWMWMWM

Therefore Statement 1 is not sufficient to answer the question.

2) Just by using Statement 2, there could be multiple scenarios where we could arrange in the x+1 men and x women in 3(5!)(7!).
Hence we cannot obtain unique value of the total number of people (that is sum of number of men and women).

However, using Statement 1 and Statement 2, the number of ways of arranging x women and x+1 men such that no two women are together are:-



(x+2)C(x) . x! . (x+1)! = 3.(5!).(7!)


By solving the above equation, we get x=5, which is the number of women.
Number of men = x+1 = 6.
Hence, total number of people = 11.

Therefore, answer is C.

Vyshak, VeritasPrepKarishma, Skywalker18, mikemcgarry, Abhishek009, msk0657, abhimahna, Bunuel, Engr2012
Can someone explain how this highlighted line be calculated?
Thanks
User avatar
mikemcgarry
User avatar
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Last visit: 06 Aug 2018
Posts: 4,474
Own Kudos:
30,882
 [2]
Given Kudos: 130
Expert
Expert reply
Posts: 4,474
Kudos: 30,882
 [2]
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
A few men and a few women are seated in a row. The number of men is one greater than the number of women. What is the total number of people?

(1) The people are seated so that no two women sit beside one another.
(2) The number of ways of arranging the seating positions of the men and women is 3(5!)(7!).
manishtank1988
Vyshak, VeritasPrepKarishma, Skywalker18, mikemcgarry, Abhishek009, msk0657, abhimahna, Bunuel, Engr2012
Can someone explain how this highlighted line be calculated?
Thanks
Dear manishtank1988,

I'm happy to respond. :-)

My friend, here's how I would think through the problem.
Statement #1: a seating constraint, but no number specific. Insufficient by itself.
Statement #2: a number given, but the number of possible seating constraints (e.g women must sit in clusters of 2 or 3 only) is mind-boggling. Insufficient by itself.

Combined: we have a clear seating constraint, which will determine a specific number of seating arrangements for each number of people, and the former will increase rapidly for each one we add to the latter. Thus, specifying a particular number of seating arrangements determines the number of people. Sufficient together.

OA = (C)

This is GMAT DS, so on principle, I would refuse to do a single calculation in answering the question. Nevertheless, you would like to see the calculation.

Again, here's how I would think about it.

Let's say there are x women. Just for the visual, I am showing the x = 5 case. These are spaced apart:

_ _ _ F _ _ _ F _ _ _ F _ _ _ F _ _ _ F _ _ _

There must be at least one man in each of the (x - 1) spaces between them. Put (x - 1) men there.

_ _ _ F _ M _ F _ M _ F _ M _ F _ M _ F _ _ _

We have two men left. Where do we put these last two men? We have (x - 1) slots between the two women as well as two on the ends, for (x + 1) total slots. We could either put both of these two extra men together, or we could put them in two different slots.
If they are together, we could put them in (x + 1). Possible outcomes:

M M _ F _ M _ F _ M _ F _ M _ F _ M _ F _ _ _

_ _ _ F _ M _ F _ M M M _ F _ M _ F _ M _ F _ _ _

_ _ _ F _ M _ F _ M _ F _ M _ F _ M _ F _ M M

If we put the two extra men in two different slots, we would have (1/2)(x+1)(x) choices. Possible outcomes:

_ M _ F _ M _ F _ M _ F _ M M _ F _ M _ F _ _ _

_ _ _ F _ M M _ F _ M _ F _ M _ F _ M M _ F _ _ _

Total number of "men slots" = (x + 1) + (1/2)(x + 1)x

Multiply that by (x+ 1)!, and then multiply this by x! for the women.

I get a different equation. I get
total number of arrangements = [(x + 1) + (1/2)(x + 1)x]*(x+ 1)!*x!

If x = 5, then this becomes

total number of arrangements = (6 + (1/2)(6)(5))*6!*5! = 21*6!*5! = 3*7*6!*5! = 3*7!*5! = 3(5!)(7!)

Does all this make sense?
Mike :-)
avatar
manishtank1988
Joined: 14 Oct 2012
Last visit: 31 Oct 2019
Posts: 112
Own Kudos:
287
 [1]
Given Kudos: 1,023
Products:
Posts: 112
Kudos: 287
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
mikemcgarry
Bunuel
A few men and a few women are seated in a row. The number of men is one greater than the number of women. What is the total number of people?

(1) The people are seated so that no two women sit beside one another.
(2) The number of ways of arranging the seating positions of the men and women is 3(5!)(7!).
manishtank1988
Vyshak, VeritasPrepKarishma, Skywalker18, mikemcgarry, Abhishek009, msk0657, abhimahna, Bunuel, Engr2012
Can someone explain how this highlighted line be calculated?
Thanks
Dear manishtank1988,

I'm happy to respond. :-)

My friend, here's how I would think through the problem.
Statement #1: a seating constraint, but no number specific. Insufficient by itself.
Statement #2: a number given, but the number of possible seating constraints (e.g women must sit in clusters of 2 or 3 only) is mind-boggling. Insufficient by itself.

Combined: we have a clear seating constraint, which will determine a specific number of seating arrangements for each number of people, and the former will increase rapidly for each one we add to the latter. Thus, specifying a particular number of seating arrangements determines the number of people. Sufficient together.

OA = (C)

This is GMAT DS, so on principle, I would refuse to do a single calculation in answering the question. Nevertheless, you would like to see the calculation.

Again, here's how I would think about it.

Let's say there are x women. Just for the visual, I am showing the x = 5 case. These are spaced apart:

_ _ _ F _ _ _ F _ _ _ F _ _ _ F _ _ _ F _ _ _

There must be at least one man in each of the (x - 1) spaces between them. Put (x - 1) men there.

_ _ _ F _ M _ F _ M _ F _ M _ F _ M _ F _ _ _

We have two men left. Where do we put these last two men? We have (x - 1) slots between the two women as well as two on the ends, for (x + 1) total slots. We could either put both of these two extra men together, or we could put them in two different slots.
If they are together, we could put them in (x + 1). Possible outcomes:

M M _ F _ M _ F _ M _ F _ M _ F _ M _ F _ _ _

_ _ _ F _ M _ F _ M M M _ F _ M _ F _ M _ F _ _ _

_ _ _ F _ M _ F _ M _ F _ M _ F _ M _ F _ M M

If we put the two extra men in two different slots, we would have (1/2)(x+1)(x) choices. Possible outcomes:
_ M _ F _ M _ F _ M _ F _ M M _ F _ M _ F _ _ _

_ _ _ F _ M M _ F _ M _ F _ M _ F _ M M _ F _ _ _

Total number of "men slots" = (x + 1) + (1/2)(x + 1)x

Multiply that by (x+ 1)!, and then multiply this by x! for the women.

I get a different equation. I get
total number of arrangements = [(x + 1) + (1/2)(x + 1)x]*(x+ 1)!*x!

If x = 5, then this becomes

total number of arrangements = (6 + (1/2)(6)(5))*6!*5! = 21*6!*5! = 3*7*6!*5! = 3*7!*5! = 3(5!)(7!)

Does all this make sense?
Mike :-)

mikemcgarry
First of all, thanks a lot for responding. I appreciate your help.
But i still have a few questions -
[Q-1] I can understand (x+1) => 1st M of the remaining 2 Ms can take (x+1) positions and 2nd M can take [(x+1)-1] = x positions. I don't understand why are we getting this (1/2)?

If we put the two extra men in two different slots, we would have (1/2)(x+1)(x) choices. Possible outcomes:
_ M _ F _ M _ F _ M _ F _ M M _ F _ M _ F _ _ _
_ _ _ F _ M M _ F _ M _ F _ M _ F _ M M _ F _ _ _


[Q-2] Also, i am assuming that we are multiplying by
(x+1)! => because finally there are (x+1) seats for men and we can arrange them in (x+1)! ways &
x! => because finally there are x seats for women and we can arrange them in x! ways
Am i correct in assuming this?
Thanks and looking forward to your reply. :P
User avatar
mikemcgarry
User avatar
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Last visit: 06 Aug 2018
Posts: 4,474
Own Kudos:
30,882
 [2]
Given Kudos: 130
Expert
Expert reply
Posts: 4,474
Kudos: 30,882
 [2]
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
manishtank1988
mikemcgarry
First of all, thanks a lot for responding. I appreciate your help.
But i still have a few questions -
[Q-1] I can understand (x+1) => 1st M of the remaining 2 Ms can take (x+1) positions and 2nd M can take [(x+1)-1] = x positions. I don't understand why are we getting this (1/2)?

If we put the two extra men in two different slots, we would have (1/2)(x+1)(x) choices. Possible outcomes:
_ M _ F _ M _ F _ M _ F _ M M _ F _ M _ F _ _ _
_ _ _ F _ M M _ F _ M _ F _ M _ F _ M M _ F _ _ _


[Q-2] Also, i am assuming that we are multiplying by
(x+1)! => because finally there are (x+1) seats for men and we can arrange them in (x+1)! ways &
x! => because finally there are x seats for women and we can arrange them in x! ways
Am i correct in assuming this?
Thanks and looking forward to your reply. :P
Dear manishtank1988,

I'm happy to respond. :-)

For Q1,
We are going to put the two men in two different slots, and there are (x + 1) slots. It's perfectly true we have (x + 1) choices for the first slot, and x choices for the second slot.

Let's say that x = 10, so we have 11 choices for the first slot and 10 choices for the second slot. Think about these two particular results:
Result #1: first choice: seventh slot; second choice: second slot
Result #2: first choice: second slot; second choice: seventh slot
These two different choices result in the same outcome, because order of picking the slots doesn't matter--all that matters is the two resultant slots.

In fact, every single result is counted twice, once where the leftmost slot was counted before the rightmost slot, and vice versa. We divide by two to eliminate the duplicates. That's why there has to be a 1/2 in the formula. In this situation, the result would be (1/2)(11)(10) = 11*5 = 55.

In fact, think about the combinations formula.
\(nCr = \tfrac{n!}{r!(n-r)!}\)

For the r = 2 case, this becomes
\(nC2 = \tfrac{n!}{2(n-2)!}\)

For n = 11, this becomes
\(nC2 = \tfrac{11!}{2(9)!} = \tfrac{11*10*9!!}{2(9)!} = \tfrac{11*10}{2} = 11*5 = 55\)

Thus even the combinations formula gives us that 1/2.

Here are two blogs you may find helpful.
GMAT Permutations and Combinations
GMAT Math: Calculating Combinations

In fact, this is really handy shortcut to know: nC2 = n(n - 1)/2

For Q2, you are 100% correct.

Does all this make sense?
Mike :-)
User avatar
mcolbert
Joined: 25 Mar 2014
Last visit: 09 Mar 2026
Posts: 12
Own Kudos:
8
 [2]
Given Kudos: 190
Posts: 12
Kudos: 8
 [2]
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Can someone please explain why the answer is not B?

We know that the total number of people is 2w+1. Therefore, the number of ways to arrange the people is (2w+1)!

(2w+1)! = 3(5!)(7!), sufficient

The key here is that the question does not say that all the men are identical and all the women are identical. Why would anyone assume that?
avatar
avikroy
Joined: 04 Jun 2010
Last visit: 26 Mar 2020
Posts: 94
Own Kudos:
Given Kudos: 264
Location: India
GMAT 1: 660 Q49 V31
GPA: 3.22
GMAT 1: 660 Q49 V31
Posts: 94
Kudos: 34
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
A few men and a few women are seated in a row. The number of men is one greater than the number of women. What is the total number of people?

(1) The people are seated so that no two women sit beside one another.
(2) The number of ways of arranging the seating positions of the men and women is 3(5!)(7!).



hello Sir,

I am having problems conceptualising the arrangement.................. Can you please help me??

the way I see it, first we can place the women ...lets say 5.... that is 5!

Then we can place atleast 1 man between 2 women............So 4 out of 6 Men can be chosen in 6C4 ways and arranged in 4! ways...

So total arrangement is 5!*6C4*4!

How to arrange the last 2 men??
avatar
Amlu95
Joined: 21 Apr 2019
Last visit: 26 Nov 2020
Posts: 30
Own Kudos:
Given Kudos: 27
Location: India
GMAT 1: 640 Q41 V37
GMAT 1: 640 Q41 V37
Posts: 30
Kudos: 15
Kudos
Add Kudos
Bookmarks
Bookmark this Post
mcolbert
Can someone please explain why the answer is not B?

We know that the total number of people is 2w+1. Therefore, the number of ways to arrange the people is (2w+1)!

(2w+1)! = 3(5!)(7!), sufficient

The key here is that the question does not say that all the men are identical and all the women are identical. Why would anyone assume that?

Its generic Ex .5 people and note it doesn't say Daniel, Mat and sandy were seated. So from this perspective, all people are identical
User avatar
srik410
Joined: 07 Oct 2024
Last visit: 18 Feb 2026
Posts: 84
Own Kudos:
Given Kudos: 246
Location: India
Concentration: Finance, Entrepreneurship
GMAT Focus 1: 685 Q90 V81 DI81 (Online)
GPA: 3.2
WE:General Management (Technology)
GMAT Focus 1: 685 Q90 V81 DI81 (Online)
Posts: 84
Kudos: 23
Kudos
Add Kudos
Bookmarks
Bookmark this Post
mcolbert
Can someone please explain why the answer is not B?

We know that the total number of people is 2w+1. Therefore, the number of ways to arrange the people is (2w+1)!

(2w+1)! = 3(5!)(7!), sufficient

The key here is that the question does not say that all the men are identical and all the women are identical. Why would anyone assume that?
This is okay to assume. In fact you have to assume everybody is different. If its same, the answer would not be C without this assumption

However this is no value of w satisfying the above equation. So this has no relevance to the standalone situation with x+1M and xW. But, with I, this makes sense and the value comes out okay
User avatar
srik410
Joined: 07 Oct 2024
Last visit: 18 Feb 2026
Posts: 84
Own Kudos:
Given Kudos: 246
Location: India
Concentration: Finance, Entrepreneurship
GMAT Focus 1: 685 Q90 V81 DI81 (Online)
GPA: 3.2
WE:General Management (Technology)
GMAT Focus 1: 685 Q90 V81 DI81 (Online)
Posts: 84
Kudos: 23
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Concise solution for C:
Assuming (1) and (2), lets see how many ways are possible to seat x+1Men and xWomen such that no two women are sitting beside each other.
Step1: Arrange all the men in the line and permutations -> (x+1)! ways
Step2: Choose x gaps from all the gaps possible with x+1 men in a line. This can be done x+2Cx ways.
Step3: Permutate all women within the chose gaps -> x!

So total = (x+1)!*(x+2)Cx*x!. Solving this we get x=5 so total=11
Moderators:
Math Expert
109818 posts
498 posts
212 posts