GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 01 Jun 2020, 09:22

### 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

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

# Find the ways in which 4 boys and 4 girls can be seated alternatively

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Intern
Status: student
Joined: 02 May 2011
Posts: 16
Location: Italy
Find the ways in which 4 boys and 4 girls can be seated alternatively  [#permalink]

### Show Tags

21 Jan 2012, 10:01
1
2
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 2 sessions

### HideShow timer Statistics

Find the ways in which 4 boys and 4 girls can be seated alternatively in a row and there is a boy J and a girl Y who can't be put in adjacent seats:

1)BGBGBGBG= 4*4*3*3*2*2*1*1 *2= 4!*4!*2 total ways
2)JY BGBGBG= 7* 3*3*2*2*1*1 = 3!*3!*7

now I guess I have to multiplied 2) by 4 because:
-JY BGBGBG
-YJ BGBGBG
-JY GBGBGB
-YJ GBGBGB

4!*4!*2 -(3!*3!*7)*4= result

do you think am I right?
Math Expert
Joined: 02 Sep 2009
Posts: 64117
Re: Find the ways in which 4 boys and 4 girls can be seated alternatively  [#permalink]

### Show Tags

21 Jan 2012, 11:37
1
3
mushyyy wrote:
Find the ways in which 4 boys and 4 girls can be seated alternatively in a row and there is a boy J and a girl Y who can't be put in adjacent seats:

1)BGBGBGBG= 4*4*3*3*2*2*1*1 *2= 4!*4!*2 total ways
2)JY BGBGBG= 7* 3*3*2*2*1*1 = 3!*3!*7

now I guess I have to multiplied 2) by 4 because:
-JY BGBGBG
-YJ BGBGBG
-JY GBGBGB
-YJ GBGBGB

4!*4!*2 -(3!*3!*7)*4= result

do you think am I right?

First of all, please post answer choices for PS questions. Also do no reword or shorten the stem.

Next, I wouldn't worry about this question at all as it's out of the scope of the GMAT. On the GMAT combination/probability questions are fairly straightforward and there is no need to spent too much time on this kind of problems.

Anyway:
There are 4 boys and 4 girls. We should sit them in a row so that boys and girls are seated alternatively and one particular boy (B') and one particular girl (G') are not in adjacent seats.

1. 4 boys and 4 girls can be seated alternately in 4!*4!*2 ways:
BGBGBGBG - 4!*4! (4 boys can be arranged in their slots in 4! ways and similarly 4 girls can be arranged in their slots in 4! ways);
GBGBGBGB - 4!*4! (the same here, we just start a row with a girl this time).

2. Let's count total # of ways B' and G' are in adjacent seats:

Consider them to be one unit {B'G'}
If we start a row with a boy: *BG*BG*BG* --> we can place {B'G'} in one of the 4 empty slots (notice that we are not violating alternately seating restriction in this case) and other 3 girls and 3 boys can be arranged in 3!*3! ways --> 4*3!*3!;
If we start a row with a girl: G*BG*BG*B --> we can place {B'G'} in one of the 3 empty slots and other 3 girls and 3 boys can be arranged in 3!*3! ways --> 3*3!*3!;
Total for the unit {B'G'}: 4*3!*3!+3*3!*3!=7*3!*3!

The same # of ways if we consider unit to be {G'B'};

Therefore total # of ways where B' and G' are in adjacent seats is 2*7*3!*3!.

3. Finally, {total # of ways where B' and G' are not in adjacent seats and boys and girls are seated alternatively} = {total # of ways where boys and girls are seated alternatively} - {total # of ways where B' and G' are in adjacent seats} = 4!*4!*2 - 2*7*3!*3! = 648.

Hope it's clear.
_________________
Intern
Status: student
Joined: 02 May 2011
Posts: 16
Location: Italy
Re: Find the ways in which 4 boys and 4 girls can be seated alternatively  [#permalink]

### Show Tags

21 Jan 2012, 19:17
Bunuel wrote:
mushyyy wrote:
Find the ways in which 4 boys and 4 girls can be seated alternatively in a row and there is a boy J and a girl Y who can't be put in adjacent seats:

1)BGBGBGBG= 4*4*3*3*2*2*1*1 *2= 4!*4!*2 total ways
2)JY BGBGBG= 7* 3*3*2*2*1*1 = 3!*3!*7

now I guess I have to multiplied 2) by 4 because:
-JY BGBGBG
-YJ BGBGBG
-JY GBGBGB
-YJ GBGBGB

4!*4!*2 -(3!*3!*7)*4= result

do you think am I right?

First of all, please post answer choices for PS questions. Also do no reword or shorten the stem.

Next, I wouldn't worry about this question at all as it's out of the scope of the GMAT. On the GMAT combination/probability questions are fairly straightforward and there is no need to spent too much time on this kind of problems.

Anyway:
There are 4 boys and 4 girls. We should sit them in a row so that boys and girls are seated alternatively and one particular boy (B') and one particular girl (G') are not in adjacent seats.

1. 4 boys and 4 girls can be seated alternately in 4!*4!*2 ways:
BGBGBGBG - 4!*4! (4 boys can be arranged in their slots in 4! ways and similarly 4 girls can be arranged in their slots in 4! ways);
GBGBGBGB - 4!*4! (the same here, we just start a row with a girl this time).

2. Let's count total # of ways B' and G' are in adjacent seats:

Consider them to be one unit {B'G'}
If we start a row with a boy: *BG*BG*BG* --> we can place {B'G'} in one of the 4 empty slots (notice that we are not violating alternately seating restriction in this case) and other 3 girls and 3 boys can be arranged in 3!*3! ways --> 4*3!*3!;
If we start a row with a girl: G*BG*BG*B --> we can place {B'G'} in one of the 3 empty slots and other 3 girls and 3 boys can be arranged in 3!*3! ways --> 3*3!*3!;
Total for the unit {B'G'}: 4*3!*3!+3*3!*3!=7*3!*3!

The same # of ways if we consider unit to be {G'B'};

Therefore total # of ways where B' and G' are in adjacent seats is 2*7*3!*3!.

3. Finally, {total # of ways where B' and G' are not in adjacent seats and boys and girls are seated alternatively} = {total # of ways where boys and girls are seated alternatively} - {total # of ways where B' and G' are in adjacent seats} = 4!*4!*2 - 2*7*3!*3! = 648.

Hope it's clear.

thanks!anyway the stem is right this one!
Non-Human User
Joined: 09 Sep 2013
Posts: 15033
Re: Find the ways in which 4 boys and 4 girls can be seated alternatively  [#permalink]

### Show Tags

04 Sep 2018, 04:12
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: Find the ways in which 4 boys and 4 girls can be seated alternatively   [#permalink] 04 Sep 2018, 04:12

# Find the ways in which 4 boys and 4 girls can be seated alternatively

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne