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Four women and three men must be seated in a row for a group photo

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Four women and three men must be seated in a row for a group photo  [#permalink]

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New post 19 Apr 2017, 06:17
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Four women and three men must be seated in a row for a group photograph. If no two men can sit next to each other, in how many different ways can the seven people be seated?

A) 240
B) 480
C) 720
D) 1440
E) 5640

*kudos for all correct solutions

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Re: Four women and three men must be seated in a row for a group photo  [#permalink]

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New post 19 Apr 2017, 16:14
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GMATPrepNow wrote:
Four women and three men must be seated in a row for a group photograph. If no two men can sit next to each other, in how many different ways can the seven people be seated?

A) 240
B) 480
C) 720
D) 1440
E) 5640

*kudos for all correct solutions


Take the task of arranging the 7 peopl and break it into stages.

Stage 1: Arrange the 4 women in a row
We can arrange n unique objects in n! ways.
So, we can arrange the 4 women in 4! ways (= 24 ways)
So, we can complete stage 1 in 24 ways

IMPORTANT: For each arrangement of 4 women, there are 5 spaces where the 3 men can be placed.
If we let W represent each woman, we can add the spaces as follows: _W_W_W_W_
So, if we place the men in 3 of the available spaces, we can ENSURE that two men are never seated together.

Let's let A, B and C represent the 3 men.

Stage 2: Place man A in an available space.
There are 5 spaces, so we can complete stage 2 in 5 ways.

Stage 3: Place man B in an available space.
There are 4 spaces remaining, so we can complete stage 3 in 4 ways.

Stage 4: Place man C in an available space.
There are 3 spaces remaining, so we can complete stage 4 in 3 ways.

By the Fundamental Counting Principle (FCP), we can complete the 4 stages (and thus seat all 7 people) in (24)(5)(4)(3) ways (= 1440 ways)

Answer:

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So be sure to learn this technique.

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Re: Four women and three men must be seated in a row for a group photo  [#permalink]

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New post 19 Apr 2017, 09:54
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try arrange women first : 4 women could get arrange 4! ways and they will create 5 gaps among themselves. Now, we have 5 gaps and 3 males could be seated on those gaps in 5p3 ways. Therefore, total arrangements would be 4!*5*4*3=1440 -> option C
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Re: Four women and three men must be seated in a row for a group photo  [#permalink]

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New post 19 Apr 2017, 08:52
GMATPrepNow wrote:
Four women and three men must be seated in a row for a group photograph. If no two men can sit next to each other, in how many different ways can the seven people be seated?

A) 240
B) 480
C) 720
D) 1440
E) 5640

*kudos for all correct solutions


Hi

_ M _ M _ M _

" No two men can sit next to each other" - we need to place woman between each man. This can be done in 4C4 ways. In this particular case we can say that they need to alternate.

We can arrange 4 women in 4! ways and 3 men in 3! ways.

Total # of sitting arrangements:

4C4 * 4! * 3! = 24 * 6 = 144

Not in the answer options?!

Am I missing something?
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Re: Four women and three men must be seated in a row for a group photo  [#permalink]

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New post 19 Apr 2017, 08:58
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vitaliyGMAT wrote:
GMATPrepNow wrote:
Four women and three men must be seated in a row for a group photograph. If no two men can sit next to each other, in how many different ways can the seven people be seated?

A) 240
B) 480
C) 720
D) 1440
E) 5640

*kudos for all correct solutions


Hi

_ M _ M _ M _

" No two men can sit next to each other" - we need to place woman between each man. This can be done in 4C4 ways. In this particular case we can say that they need to alternate.

We can arrange 4 women in 4! ways and 3 men in 3! ways.

Total # of sitting arrangements:

4C4 * 4! * 3! = 24 * 6 = 144

Not in the answer options?!

Am I missing something?


Your solution only allows for one configuration: W M W M W M W
However, we can also have W W M W M W M or W M W W M W M, etc.

Cheers,
Brent
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Re: Four women and three men must be seated in a row for a group photo  [#permalink]

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New post 19 Apr 2017, 09:11
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GMATPrepNow wrote:
vitaliyGMAT wrote:
GMATPrepNow wrote:
Four women and three men must be seated in a row for a group photograph. If no two men can sit next to each other, in how many different ways can the seven people be seated?

A) 240
B) 480
C) 720
D) 1440
E) 5640

*kudos for all correct solutions


Hi

_ M _ M _ M _

" No two men can sit next to each other" - we need to place woman between each man. This can be done in 4C4 ways. In this particular case we can say that they need to alternate.

We can arrange 4 women in 4! ways and 3 men in 3! ways.

Total # of sitting arrangements:

4C4 * 4! * 3! = 24 * 6 = 144

Not in the answer options?!

Am I missing something?


Your solution only allows for one configuration: W M W M W M W
However, we can also have W W M W M W M or W M W W M W M, etc.

Cheers,
Brent


Agrrrr...

I've chosen wrong gaps! Thanks.

We should choose seats between women.

Final solution:

5C3*4!*3! = 1440.

Good question.
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Re: Four women and three men must be seated in a row for a group photo  [#permalink]

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New post 19 Apr 2017, 09:55
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try arrange women first : 4 women could get arrange 4! ways and they will create 5 gaps among themselves. Now, we have 5 gaps and 3 males could be seated on those gaps in 5p3 ways. Therefore, total arrangements would be 4!*5*4*3=1440 -> option C
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Re: Four women and three men must be seated in a row for a group photo  [#permalink]

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New post 19 Apr 2017, 17:22
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GMATPrepNow wrote:
Four women and three men must be seated in a row for a group photograph. If no two men can sit next to each other, in how many different ways can the seven people be seated?

A) 240
B) 480
C) 720
D) 1440
E) 5640

*kudos for all correct solutions



no two men can sit together so, we have

_W_W_W_W_


SO Women can sit in 4! ways
3 men have 5 spaces to sit so: (5C3) * 3!(sort among themselves)

TOTAL = 4!* 3! * 5C2 = 1440
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Re: Four women and three men must be seated in a row for a group photo  [#permalink]

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New post 20 Apr 2017, 02:36
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_W_W_W_W_

4 women = 4! = 24
3 men = 3! = 6

24*6 = 144

now out of 5 spaces any 3 coz 3 men

5c3 = 5c2 = 10


144*10 = 1440
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Re: Four women and three men must be seated in a row for a group photo  [#permalink]

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New post 27 Apr 2017, 17:11
Can someone help me understand the assumption that there are 5 chairs available after the 4 women have been seated? Seems to me that this is an insufficiently detailed question prompt. Wouldn't a more logical assumption be that there are 3 remaining chairs after the 4 women have been seated for a total of 7 chairs - the same as the number of people?

Thanks.
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Four women and three men must be seated in a row for a group photo  [#permalink]

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New post 27 Apr 2017, 23:08
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Shazriki wrote:
Can someone help me understand the assumption that there are 5 chairs available after the 4 women have been seated? Seems to me that this is an insufficiently detailed question prompt. Wouldn't a more logical assumption be that there are 3 remaining chairs after the 4 women have been seated for a total of 7 chairs - the same as the number of people?

Thanks.


Hi there ,
Kindly visualise this W_W _W_W

let's name these 3 spaces 1,2,3 from left to right
i.e w 1 w 2 w 3 w

but what if the person chooses to sit like this
1ww2w3
1w2ww3w
though it satisfies the initial condition , what you have assumed fails in such cases

hence to avoid this confusion , we arrange 4 women _W_W_W_W_

please note that the _ represents a space and 3 spaces out of these 5 spaces can be chosen. w1w2W3w is not the only combination , there would be other possibilities as shown above

lets say you are seated and there is a possibility that a person can sit at either side of you viz. Left/ right
i hope this is clear

kudos if this helps
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Re: Four women and three men must be seated in a row for a group photo  [#permalink]

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New post 17 May 2017, 01:41
GMATPrepNow wrote:
Four women and three men must be seated in a row for a group photograph. If no two men can sit next to each other, in how many different ways can the seven people be seated?

A) 240
B) 480
C) 720
D) 1440
E) 5640

*kudos for all correct solutions


Let three men be A,B,C

total number of ways =
4 w & A = 5! x4
4 w & B =5! x 4
4 w & C = 5! x 4 = 3(5!x4)= 1440

multiplied by 4 : becaause the man can sit with any of the 4 women.

is my approach correct?
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Re: Four women and three men must be seated in a row for a group photo  [#permalink]

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New post 17 May 2017, 02:28
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first let sit 4 women.

This can be done in 4! ways = 4*3*2*1=24 ways
among 4 women , there are 5 spaces
3 men can sit in these 5 sits in 5C3* 3!ways = 10 *3*2*1=60 ways
total ways =24*60=1440
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Re: Four women and three men must be seated in a row for a group photo  [#permalink]

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New post 17 May 2017, 05:23
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hanyhamdani wrote:
GMATPrepNow wrote:
Four women and three men must be seated in a row for a group photograph. If no two men can sit next to each other, in how many different ways can the seven people be seated?

A) 240
B) 480
C) 720
D) 1440
E) 5640

*kudos for all correct solutions


Let three men be A,B,C

total number of ways =
4 w & A = 5! x 4
4 w & B =5! x 4
4 w & C = 5! x 4 = 3(5!x4)= 1440

multiplied by 4 : becaause the man can sit with any of the 4 women.

is my approach correct?


It's hard to tell whether your approach is correct.
What does "4 w & A = 5! x 4" represent?

Cheers,
Brent
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Re: Four women and three men must be seated in a row for a group photo  [#permalink]

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New post 18 Sep 2018, 16:32
Someone please explain where I have gone wrong -

1) Unconstrained way of arranging 4W & 3M -> 7!
2) Now, to formulate scenarios when no men can sit together, I have grouped them. Hence now we have 5 entities 4W & 1M (group). Also the men can be arranged 3! ways.

3) So the final answer is -> 7! - 5!3!
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Four women and three men must be seated in a row for a group photo  [#permalink]

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New post 05 Jun 2019, 21:25
GMATPrepNow wrote:
GMATPrepNow wrote:
Four women and three men must be seated in a row for a group photograph. If no two men can sit next to each other, in how many different ways can the seven people be seated?

A) 240
B) 480
C) 720
D) 1440
E) 5640

*kudos for all correct solutions


Take the task of arranging the 7 peopl and break it into stages.

Stage 1: Arrange the 4 women in a row
We can arrange n unique objects in n! ways.
So, we can arrange the 4 women in 4! ways (= 24 ways)
So, we can complete stage 1 in 24 ways

IMPORTANT: For each arrangement of 4 women, there are 5 spaces where the 3 men can be placed.
If we let W represent each woman, we can add the spaces as follows: _W_W_W_W_
So, if we place the men in 3 of the available spaces, we can ENSURE that two men are never seated together.

Let's let A, B and C represent the 3 men.

Stage 2: Place man A in an available space.
There are 5 spaces, so we can complete stage 2 in 5 ways.

Stage 3: Place man B in an available space.
There are 4 spaces remaining, so we can complete stage 3 in 4 ways.

Stage 4: Place man C in an available space.
There are 3 spaces remaining, so we can complete stage 4 in 3 ways.

By the Fundamental Counting Principle (FCP), we can complete the 4 stages (and thus seat all 7 people) in (24)(5)(4)(3) ways (= 1440 ways)

Answer:

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So be sure to learn this technique.

RELATED VIDEOS





Hi GMATPrepNow

So what i did was i did the total- not possible. Total being 7!=5040

Not possible : 3 Men M1,M2,M3

1st condition M1 and M2 not sit together : 1440
2nd condition M2 and M3:1440
3rd condition M3 and M1 :1440

Subtracting i got 720. Please tell me where i got my logic wrong
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Re: Four women and three men must be seated in a row for a group photo  [#permalink]

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New post 06 Jun 2019, 02:57
Consider the arrangement *W*W*W*W*

The three men can be placed in any 3 of the 5 *s to ensure that no 2 men sit next to each other

No of ways of choosing 3 * out of 5 = 5C3

The three men can arrange themselves in 3! ways and similarly the four women can arrange themselves in 4! ways

Therefore, the total number of ways in which the seven can sit with no two men next to each other = 5C3*3!*4! = 10*6*24 = 1440

Answer is (D)

Hit Kudos if this helped!
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Re: Four women and three men must be seated in a row for a group photo  [#permalink]

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New post 06 Jun 2019, 06:03
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AlN wrote:
Hi GMATPrepNow

So what i did was i did the total- not possible. Total being 7!=5040

Not possible : 3 Men M1,M2,M3

1st condition M1 and M2 not sit together : 1440
2nd condition M2 and M3:1440
3rd condition M3 and M1 :1440

Subtracting i got 720. Please tell me where i got my logic wrong


How are you calculating the 3 values (of 1440)?
Also, if 5040 = the total number of arrangements
And if 3(1440) = the number of arrangements in which the men are NOT together, then 5040 - 3(1440) = the number of arrangements in which the men ARE together (and the question says the men CANNOT be together)

Cheers,
Brent
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Re: Four women and three men must be seated in a row for a group photo   [#permalink] 06 Jun 2019, 06:03
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