In how many ways can 3 girls and 4 boys be seated in a row such that

Question

In how many ways can 3 girls and 4 boys be seated in a row such that no two girls are seated next to each other?

A. 144
B. 480
C. 720
D. 960
E. 1440

gmatophobia · Answer

_B _ B _ B _ B _

Four boys (denoted by B) can be seated in 4! ways.

As no two girls sit together, the three girls can be seated in the space represented by the dashes '_'.

Out of the five available dashes, we can select 3 in  ^5C_3  ways. Once selected, the three girls can be seated in 3! ways.

The number of possible arrangements =  4! * ^5C_3 * 3! = 1440

Option E

missionmba2025 · Answer

Assume that four boys are seated in the four dashes

_ _ _ _

The seating can be done in 4! ways.

In the gaps, shown by 'x', we can seat the girls.

x _ x _ x_ x_ x

Three places of the five places ('X's) can be selected in  C^5_3  ways, and the girls can be arranged in 3! ways.

Total possibilities =  4! * C^5_3 * 3!  = 1440 ways

IMO E

8.toastcrunch · Answer

To find the number of ways to seat the 3 girls and 4 boys in a row such that no two girls are seated next to each other, we can use the concept of permutations.

First, we can arrange the boys in a row. There are 4 boys, so there are 4! = 4 x 3 x 2 x 1 = 24 ways to arrange them.

Next, we need to consider the positions where the girls can be seated. We have 5 possible positions: before the first boy, between the first and second boy, between the second and third boy, between the third and fourth boy, and after the fourth boy.

Out of these 5 positions, we need to select 3 positions for the girls to be seated. We can choose these positions in C(5, 3) = 5! / (3! x (5-3)!) = 5! / (3! x 2!) = 10 ways.

Once the positions for the girls are selected, we can arrange the girls among themselves in those positions. There are 3 girls, so there are 3! = 3 x 2 x 1 = 6 ways to arrange them.

Therefore, the total number of ways to seat the girls and boys is  24 x 10 x 6 = 1440

The correct option is E

IshanSaini · Answer

I am new to PnC and I have a question here :
I understand that if I arrange the girls first fixing them in three places we have 3! shuffled ways for them
but now I have
_ G _ G _ G _ 
where _ indicates the number of positions where I can place boys meaning that there are 4C4 i.e. 1 such arrangement possible and 4! shuffled ways.
Here lies my doubt.

So can the answer be 4! * 3! * 4C4 = 144.

BUT

As explained by the posts above if I place Boys first then I have 
_ B _ B _ B _ B _ indicating 4! * 3! * 5C3 cases or 1440 ways.

What am i missing here ?

walterwhite756 · Answer

Nice question there,  IshanSaini .

See in your first method you are missing out of 2 extra possible places of Girls. In "   B1  G  B2  G  B3  G  B4   " A girl can sit ahead of B1 and after B4 as well. Out of  5  possible places, you are counting only  3  when you fix girl's position first.

Instead, what you should always do is fix places of  NON-CONSTRAINT  items first. Here we don't have any condition on placement of Boys and so first place them and then apply the condition that girls should sit apart from each other. Don't linger much on lingo, instead focus on the way these questions can be solved.

Hope it helped. Feel free to ping me if any doubt persists.

IshanSaini · Answer

Thanks For the quick reply  walterwhite756 .
Yes, i see the point you are trying to make. Seems fair and i understand what was the error.

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In how many ways can 3 girls and 4 boys be seated in a row such that

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In how many ways can 3 girls and 4 boys be seated in a row such that

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