A group of four boys and three girls is to be seated in a row. how man

Question

A group of four boys and three girls is to be seated in a row. how many such arrangements are possible where no girls sit together ?

A. 144
B. 288
C. 576
D. 720
E. 1440

Source:- Experts global.

GMATinsight · Accepted Answer

Between every two BOYS there must be A MAXIMUM OF one GIRL hence we the the arrangements will look as follows

- B - B - B - B -

But since there are only 3 girls and 5 places available for them (shown by dashes above) so we need to select those 3 places for the girls out of 5

Selection of 3 places out of 5 for the girls = 5C3
Arrangement of the girls on 3 selected places = 3!
Arrangement of the boys on 4 places = 4!

Total Favourable outcomes = 5C3*3!*4! = 60*24 = 1440

Answer: option E

GMATBusters · Answer

The 4 boys can be seated in 4! Ways.
The 3 girls can be seated between boys .
We have to selected 3 positions from 5 available.
5C3.
Ordering =3!.

Total number of ways = 4!*5C3*3! = 1440. Answer E

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SalmanDard · Answer

I am trying to understand why are you assuming 9 spaces to be filled? are we not supposed to fill 7 spaces (4 boys and 3 girls)? Kindly help me understand!
"B-B-B-B"

GMATBusters · Answer

I would be happy to explain.

Since the condition is that no 2 girls should sit together, we should ensure that there would be at least one boy between any 2 girls.
Now the 4 boys have already been placed in 4! ways.

After that we need to find the slots which can be filled by girls so that no 2 girls are together,

- B - B - B - B -

If you observe carefully,  if the girls are made to sit on the 5 blanks seat represented by -, then no 2 girls would be together.

So as we have 3 girls , we need to select 3 blanks seat out of 5 blanks as mentioned above, which can be done in 5C3 ways.
After the selection of 3 blanks seats, the girls can be arranged in 3! ways.

Hence the total number of ways would be 4!*5C3*3! = 1440.

I hope it is clear now, feel free to ask further in case of any doubts.

SalmanDard · Answer

BGBGBGB <- this fulfills 4 boys and 3 girls and no girls sit together. So 4!x3!=144 which is A???

GMATBusters · Answer

Hi I

You are right , this fulfills the criteria, but this covers some of the favourable cases, not all of them.
We need all the favourable cases, which can be found as explained above.
Feel free to ask again if not fully satisfied.

Regards

aviejay · Answer

I think this question can be solved even using jst permutation. However I am not able to arrive at the answerusing jst permutations. Can someone please help me?

The concept that I used is 7!-(all combinations wth 2 girls together) - (all combinations wth 3 girls together)

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@s · Answer

Hi Let me try to explain this, The statement should be 7! - = 5040 - = 1440 Total number of ways in which 7 students can sit is 7!. Total number of ways in which 2 girls sit together - 6! 2! 2! . Total number of ways in which 3 girls sit together - 5! 3! Hope I was able to clear your doubt.

Lowkya · Answer

There are four boys and three girls.
Condition: No girls should sit together.
In order to make sure that the condition is satisfied, we order the boys first and then the girls in the gaps between the boys. 
As the number of boys here is 4. They can be arranged in 4! ways.
  __B__B__B__B__  
We have 5 gaps between the boys. ( In general, gaps will be number of things + 1 )
We need to first choose 3 gaps for the girls, this can be done in 5C3 ways.
Then arrange the girls, which can be done in 3! ways.
Total: 4! X 5C3 X 3! ( We need to multiply the events here )
 Total = 1440. 
 Option E

Shef08 · Answer

Can someone explain me where did I go wrong?

Ways in which 7 people can be arranged: 7!.... (a) 
Ways in which all girls are arranged together: 7P5...... (b) 
(as all girls are treated as 1) 
So a-b: 7!- 7P5

firas92 · Answer

You have negated only the cases where all 3 girls are seated together.

There are still other scenarios where two girls can be seated together. For example BGGBBBG, GBGGBBB, GGBBGBB, etc. These are also not allowed as per our question. No two girls can be seated together.

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Shef08 · Answer

Thank you Firas, I know my mistake now!!!

kc2025 · Answer

but there can also be a gap to two seats between the girls correct?

Bunuel · Answer

So if I understand you correctly, you are basically saying that “no two girls sitting together” could mean they could be separated not only by a boy but also by an empty seat. The question should have mentioned explicitly that there are exactly 7 seats for the 7 people. With that condition stated, there is no possibility of empty seats, and the only way to ensure no two girls sit together is as shown above.

kc2025 · Answer

no, I'M saying they could be separated by 2 boys not just 1 boy

egmat · Answer

kc2025  - Looking at your doubt, I can see there's a key misunderstanding about what the gaps method accomplishes.

Understanding Your Confusion

Your concern that "there can also be a gap to two seats between the girls" shows you're thinking about this correctly - girls  can indeed  be separated by 2 boys (or even 3 boys)! The beautiful part is that  the gaps method already accounts for this .

What "No Girls Sit Together" Really Means

This constraint means:  No two girls should be adjacent/next to each other  in the row. It does  not  mean girls must be separated by exactly one boy.

How the Gaps Method Works

When we arrange  4  boys first, we get:
__B__B__B__B__

These  5  gaps represent  potential positions  where girls can sit, not mandatory separators.

Key Insight:  When we choose  3  different gaps from these  5  gaps and place one girl in each:
 
 Some girls will be separated by  1  boy 
 Some girls might be separated by  2  boys 
 Some girls might be separated by  3  boys

Example Arrangements

Let's say boys are B1, B2, B3, B4 and girls are G1, G2, G3:

If we choose gaps 1, 3, and 5: 
G1 B1 B2 G2 B3 B4 G3
→ G1 and G2 are separated by  2  boys
→ G2 and G3 are separated by  2  boys

If we choose gaps 1, 2, and 5: 
G1 B1 G2 B2 B3 B4 G3
→ G1 and G2 are separated by  1  boy
→ G2 and G3 are separated by  3  boys

Why This Method Is Complete

The method counts  ALL possible valid arrangements  where no girls are adjacent, regardless of whether they're separated by  1 ,  2 , or  3  boys. By choosing  3  different gaps from  5 , we guarantee girls won't be next to each other.

Common Misconception 
Students often think we need to separately count cases where girls are separated by different numbers of boys. But the gaps method elegantly handles all cases simultaneously!

The Calculation Remains: 
 4! 	imes {5 \choose 3} 	imes 3! = 24 	imes 10 	imes 6 = 1440

Answer:  E

Bunuel · Answer

Oh, I see. In this case the solutions above still hold. When we select the 3 empty spaces shown by the dashes in the pattern - B - B - B - B - , the girls can be placed in any 3 of those 5 slots. Notice that if we pick, for example, the first, second, and last slots, the arrangement could look like G B G B B B G. In this setup, some girls are separated by more than one boy. So the solution already accounts for cases where the girls are separated by more than one boy, not just by exactly one.

A group of four boys and three girls is to be seated in a row. how man

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