In how many different ways can 3 boys and 3 girls be seated in a row

Since 3 people will be on each side, and no two boys/two girls can sit together, on one side you will have Boy-Girl-Boy arrangement and on the other side you will have Girl-Boy-Girl arrangement.
From the 3 boys, choose 2 in 3C2 ways and a girl in 3C1 ways. Now you have split the boys and girls into BGB and GBG.
For BGB, pick a side (either chairs or stools) in 2 ways.
The girl takes the center seat and the 2 boys can be arranged around the girl in 2! ways.
On the other side, the boy sits in the center and the girls are arranged on his two sides in 2! ways.

Total arrangements = 3C2 * 3C1 * 2 * 2! * 2! = 72 ways.

Answer (C)

3!*3!*2! = 6*6*2 = 72

Answer C

Lets consider the group of all boys as one entity and group of girls as other entity

Now we have two entities to arrange which is possible in 2! ways

The group of boys within itself can be rearranged in 3! ways and
The group of Girls within itself can be rearranged in 3! ways

Total arrangements = 2!*3!*3! = 2*6*6 = 72

Answer: Option C

Bunch up the boys together and the girls together. Now we have to arrange 2 groups which we can do in 2! ways.

The boys within themselves can be arranged in 3! ways and the girls within themselves can be arranged in 3! ways.

Total arrangements = 2 * 3! * 3! = 72

Answer (C)

Condition: girls not separated..and boys are not separated

Count them as 1&1. Now we have 1boys n 1girl..

Arrangements =2!
And arrangements inside 3 girls & inside 3boys is=3! & 3!

Total=2!*3!*3! = 72 option C

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We need to determine how many ways to arrange 3 boys and 3 girls in a row of six chairs when the boys must be seated together and the girls must be seated together. Since order matters, we have a permutation problem.

We can arrange the boys and girls as follows:

[B1)-B2)-B(3)] - [G(1)-G(2)-G(3)]

(Note: Since the boys must sit together and the girls must sit together, we have included the boys in one bracket and the girls in one bracket.)

Since we have 2 brackets, we can arrange those 2 brackets in 2! or 2 ways. However, we must account for the individual arrangements of the boys and the girls. The boys can be arranged in 3! or 6 ways and the girls can be arranged in 3! or 6 ways. Thus, the group can be arranged in 2 x 6 x 6 = 72 ways.

Answer: C

Hi Al,

There are a number of ways to do the "math in this question", but here's a way that you might find easy:

The fact that we have 6 chairs in a row means that we'll likely be doing a permutation. We're given the specific rule that the seating will have to be either GGGBBB or BBBGGG.

The first chair can have either a boy or a girl:

6 _ _ _ _ _

Whether it's a boy or a girl, the next two chairs must be the same gender:

6 2 1 _ _ _

Then the next 3 chairs must be the other gender:

6 2 1 3 2 1

Now multiply:

6x2x1x3x2x1 = 72 possibilities.

Final Answer:
GMAT assassins aren't born, they're made,
Rich

ScottTargetTestPrep EMPOWERgmatRichC KarishmaB Should not the answer be 144. I multiplied 72 by 2 since person sitting on a chair and sitting on a stool is totally different arrangement. Please let me know where my assumption is wrong

Hi naikabhishekjanu,

Each of the explanations in this thread HAS accounted for the fact that there are chairs on one side of the table and stools (meaning a different type of seat) on the other). Here are some additional details that might help you understand the math a bit better:

Since there are chairs on one side of the table and stools on the other side of the table, we could have 2 boys on the chairs-side and 1 on the stools-side OR 1 boy on the chairs-side and 2 on the stools-side. Thus, we have to account for 2 possible seating arrangements:

BGB
GBG

and

GBG
BGB

From here, we can calculate two separate permutations (one for each seating option) to get to the answer. For the first calculation, let's assume that two boys are sitting on the chairs-side:

Moving from left-to-right (with the chairs-side and then through the stools-side)....
For the first "spot", there are 3 different boys to choose from
For the second "spot", there are 3 different girls to choose from
For the third "spot", there then 2 different boys to choose from
For the fourth "spot", there are then 2 different girls to choose form
For the fifth and sixth "spots", we have the 1 boy and 1 girl that are left

(3)(3)(2)(2)(1)(1) = 36 possible seating arrangements when 2 boys are on the chairs-side.

Since the math would essentially be identical if we had 2 girls on the chairs-side instead, we can multiply this total by 2...

(36)(2) = 72

Thus, there are 72 ways to arrange the 6 people as the prompt describes the seating.

GMAT assassins aren't born, they're made,
Rich

naikabhishekjanu
It's already accounted for in 72. Consider the highlighted.

EMPOWERgmatRichC KarishmaB Thanks for the explaination.

EMPOWERgmatRichC What if question were like this one: there are 3 boys and 3 girls, how many combinations are there if no boy should be between any two girls

Hi dave13,

In that situation, since we're looking for a really specific situation that we do NOT want to include, you would likely find it easiest to calculate all of the ways for THAT situation to happen - and then subtract that from the total possible ways to sit 6 people. I'll give you a couple of hints so that you can complete this on your own:

-How different ways could you sit the 6 people if there were no restrictions?

-Since the one situation that we do NOT want is a GBG on either side of the table, how many different ways could we have...?

GBG
XXX

XXX
GBG

GMAT assassins aren't born, they're made,
Rich

Contact Rich at: [email protected]

hi EMPOWERgmatRichC

thanks!

ok let me try

Without restrictions 3 girls and 3 boys can be arranged in 6 / 3! 3! = dividing by 3! 3! cause we have BBBGGG

When we have GBG then (from left to right) there are 6 ways to place a girl, 6 ways to boy and 5 ways to place a girl so 6*6*5 = 180


now Total # of combinations MINUS undesirable = 20 - 180 =? oops maybe i shouldnt divide by 3!3! but i see BBBGGG i.e. three identical elements per type) pls advice advice ?

Another question, what if we remove "table" and place 3 boys and 3 girls, in one row such that no boy should be between any two girls ?

would the solution to such question be like this one

BBBGGG
GGGBBB
BGGGBB
BBGGGB

so 4 combinations total ?

0r...

1. BBBGGG
2. GGGBBB
3.BGGGBB
4.BBGGGB

case 1) 3 *2*1*3*2*1 = 36 ( from left to right 3 boys to chose from *2 boys to choose from etc*1*3*2*1 = 36)

case 2) 3 *2*1*3*2*1 = 36

case 3) 3*3*2*1*2*1 = 36

case 4) same logic here

pls advice.

KarishmaB can you pls help with the two questions below....tried to solve these ones but cant find a correct answer

Question 1. at a rectangular table there are 3 boys and 3 girls, how many arrangements are there if no boy should be between any two girls?

Question 2. how many arrangements are there to place 3 boys and 3 girls in one row, if no boy should be between any two girls?

Hi dave13,

There were a couple of errors in how you attempted to answer your initial question. First, in the original prompt, there were chairs on one side of the table and stools on the other side (and this impacts the permutation involved). Second, there are 3 boys and 3 girls, meaning that there are NOT '6 ways to place a girl', etc. (re: a Girl must be placed in the "girl spot" and a Boy must be placed in the "boy spot"). If you go back and rework your calculations, then what result would you get?

In the second question (in which the three girls essentially had to sit 'side-by-side', your approach is correct. Each of the 4 options would have 3!(3!) iterations - so the total possible way to seat the 6 people would be 4(3!)(3!) = 144.

GMAT assassins aren't born, they're made,
Rich

Contact Rich at: [email protected]

EMPOWERgmatRichC
hey Rich , thanks!

ok let me try again solution to question one.

Total number of ways to arrange 3 boys and 3 girls (without any restrictions) is 6! = 720

number of ways when BGB = ( there are 3 boys to choose from for the first place, 3 girls to choose from for the second place, 2 boys to choose for the third place,

on the other side of the table (GBG) for the fourth place 2 girls to choose from and for the fifth and and sixth places 1 girl and 1 boy)

So, 3*3*2*2*1*1 = 36

now that I have calculated arrangements on both sides of table BGB and GBG there is also another arrangement GBG and BGB

so 36*2 = 72 not sure though

So 720 - 72 = 648 is it correct answer ?

So when arranging some numbers of boys and girls should i always treat all boys as different elements ? i mean B1 B2 B3 etc same logic applies for girls G1, G2, G3

however you seem to emphasize difference between chairs and stools in your comment below


what if on both sides of table there were chairs ? how would it impact solution ?

Hi dave13,

The overall 'intent' of this question (along with your 'variation'), is that there aren't supposed to be any 'duplicate' arrangements (and this is why the original prompt tells us that one side of the table has chairs while the other side has stools). Without that distinction, we could conceivably have a situation in which a rotated-table (re: rotated 180 degrees) can occur and the following two orientations would be considered the SAME (and not different):

ABC
DEF

FED
CBA

To keep the original intention of the prompt and your follow-up question, it is correct that there would be (6)(5)(4)(3)(2)(1) = 720 possible ways to sit 6 people at the table. You specifically asked to eliminate all of the options that include one side of the table having two boys with a girl in-between. With a total of 3 girls and 3 boys, if you DO have a BGB on one side, then that would leave 2 girls and 1 boy on the OTHER side (meaning that there would be NO possibility of a BGB on THAT side - and those remaining 3 people could sit in any of the remaining 3 spots).

Thus, the number of BGB possibilities in...

BGB
XXX

would be (3)(3)(2)(3)(2)(1) since the last 3 people on the XXX side of the table could be arranged in any of 6 arrangements). Thus there are 108 possible options here.

With the BGB happening on the other side...

XXX
BGB

we'd have another 108 options. Thus, the total number of ways that we have to subtract is 108+108 = 216 and the answer to your question is...
720 - 216 = 504.

GMAT assassins aren't born, they're made,
Rich

Contact Rich at: [email protected]