Problem Solving

There are 7 boys and 3 girls in a class. In how many ways can they be seated in a row so that no two of the 3 girls are next to each other?
a) 336 b)56* 7! c)168*6! d)336*7! e)672 *7!

Ways to seat all 10 people = 10!
Ways that the girls could be together= 8!
Ways to arrange 3 girls = 3!

Total ways to seat without girls sitting next to each other =

10!- 8!3! = 8!((10 * 9)- ( 1 * 6))= 8!(90-6)= 8!(84) = 7!(8 * 84)= 7!*672

Here is my analysis.

10! = 9!3C2(two grils in one set and one gril individual) - 8!3(all girls treated as one unit) + x (set in which no two girls sit next to each other)...

So i got x = 39*8!.

Please correct my analysis.


Thanks a lot.

Hi Rahul,
Could you please correct my analysis ....

Thanks in advance...

muralithe1 wrote:Hi Rahul,
Could you please correct my analysis ....

Thanks in advance...

I was asked by PM to discuss how you could approach this problem using the methodology that I used in the following thread: https://www.beatthegmat.com/permuatation ... tml#283495. Here goes:

Girls = ABC
Boys = D, E, F, G, H, I, J

Good arrangements = total possible arrangements - bad arrangements

Total possible arrangements = 10!

Bad arrangements:

Number of ways to arrange 3 girls together = 3!*8! = 6*8!

Number of ways to arrange AB, C, D, E, F, G, H, I, J = 9!
Number of ways to arrange ABC, D, E, F, G, H, I, J and CAB, D, E, F, G, H, I, J = 2*8!
So number of arrangements with AB together but not next to C = 9! - 2*8! = 8!*(9-2) = 7*8!
Since AB can be replaced with BA, AC, CA, BC, and CB -- for a total of 6 variations -- number of arrangements with 2 girls together but not next to 3rd girl = 6*(7 *8!) = 42*8!

So total bad arrangements = 6*8! + 42*8! = 48*8!

Good arrangements = 10! - 48*8! = 8!(90-48) = 42*8! = 336*7!.

For this problem, Rahul's method is much more efficient!

GMATGuruNY wrote:

muralithe1 wrote:Hi Rahul,
Could you please correct my analysis ....

Thanks in advance...

I was asked by PM to discuss how you could approach this problem using the methodology that I used in the following thread: https://www.beatthegmat.com/permuatation ... tml#283495. Here goes:

Girls = ABC
Boys = D, E, F, G, H, I, J

Good arrangements = total possible arrangements - bad arrangements

Total possible arrangements = 10!

Bad arrangements:

Number of ways to arrange 3 girls together = 3!*8! = 6*8!

Number of ways to arrange AB, C, D, E, F, G, H, I, J = 9!
Number of ways to arrange ABC, D, E, F, G, H, I, J and CAB, D, E, F, G, H, I, J = 2*8!
So number of arrangements with AB together but not next to C = 9! - 2*8! = 8!*(9-2) = 7*8!
Since AB can be replaced with BA, AC, CA, BC, and CB -- for a total of 6 variations -- number of arrangements with 2 girls together but not next to 3rd girl = 6*(7 *8!) = 42*8!

So total bad arrangements = 6*8! + 42*8! = 48*8!

Good arrangements = 10! - 48*8! = 8!(90-48) = 42*8! = 336*7!.

For this problem, Rahul's method is much more efficient!

thanks guru,but i did n't understood...can u explain me little bit clear.......

so then again what is the answer, i see an answer of 336 7! and 672 7!

mainhoon wrote:so then again what is the answer, i see an answer of 336 7! and 672 7!

336 * 7!

Check Rahul's post.

thanks guru,but i did n't understood...can u explain me little bit clear.......

Let's break it down step by step.

Girls = ABC
Boys = D, E, F, G, H, I, J

Good arrangements = total possible arrangements - bad arrangements

Total possible arrangements = 10! (There are 10! ways to arrange 10 elements.)

From the 10! total arrangements, we need to subtract the bad arrangements. There are two types of bad arrangements in this problem:

1) The 3 girls ABC all sit together
2) 2 girls sit together but not next to the 3rd girl

Number of ways to arrange 3 girls together:
If we think of the 3 girls as block ABC, we're arranging 8 elements (block ABC and the 7 boys). Number of ways to arrange 8 elements = 8!
ABC itself can be arranged in different orders. Number of ways to arrange ABC = 3!
Multiplying, total arrangements with the 3 girls together = 3!*8! = 6*8!

Number of ways to arrange 2 girls together but not next to the 3rd girl:
Let's start with girls AB.
Number of ways to arrange AB and the 8 other children = 9! (There are 9! ways to arrange 9 elements.)
We need to subtract all the arrangements that include ABC together and CAB together because we counted these above when we counted the number of ways to arrange all 3 girls together.
Number of ways to arrange ABC and the 7 other children is 8!
Number of ways to arrange CAB and the 7 other children is 8!
Subtracting, the number of arrangements with AB together but not next to C = 9!- 8!-8!= 8!(9-1-1)=7*8!
Since AB can be reversed to BA, the number of arrangements with BA together but not next to C = 7*8!
So bad arrangements with AB or BA next to each other but not next to C = 7*8! + 7*8! = 14*8!
Each pair of girls will give us another 14*8! bad arrangements. From ABC we can make 3*2/1*2 = 3 pairs.
So total arrangements with 2 girls together but not next to the 3rd girl = 3*(14*8!)=42*8!

So total bad arrangements = 6*8! + 42*8! = 48*8!

Good arrangements = Total arrangements - Bad arrangements = 10! - 48*8! = 8!(90-48) = 42*8! = 336*7!.

Hope this helps!

Good = Total - Bad can be an efficient way to solve many problems. For the problem above, however, determining the number of bad arrangements is complex. I suggest that you use Rahul's method, which is much more efficient!