There are 3 different cars available to transport 3 girls and 5 boys

The kids can be distributed among the 3 cars:

3 3 and 2

Since the question asks for arrangements with 2 or 3 girls in a car, we'll be adding those two scenarios.

Let's start with 2 girls in a car and no boys.

Two girls can be selected;

3!/2! = 3 ways

A car can be selected 3 ways.

So 3*3 = 9 ways

Now we have 6 people remaining to be distributed among 2 cars.

6!/3!3! = 20 are the ways to select 3 people from 6. This counts the leftover 3 as well, which isn't double counting since the 2 vehicles are different.

So 9*20 = 180

Now let's try 2 girls and 1 boy in a car.

Two girls can be selected 3 ways as above. 1 boy can be selected 5 ways and 1 car can be selected 3 ways, a total of

45 ways

The car with 2 people can be selected 2 ways. The 2 people to go in the car can be selected

5!/2!3! = 10 ways for a total of 10*2 = 20 ways.

Total ways: 45*20 = 900

Finally, let's put 3 girls in 1 car.

3 girls can be selected 1 way. A car can be selected 3 ways, for a total of

3 ways

The car with 2 people can be selected 2 ways. 2 people can be selected from 5

5!/2!3! = 10 ways

So a total of 3*2*10 = 60 ways


Totaling all of these:

180+900+60 = 1140

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How I solved this (in case the logic helps someone!)

(1) What is required? We need either 2 or 3 girls in one of the 3 cars. What are the possible scenarios? 

a) 1 car with 3 girls (GGG), then, the other two cars have only boys (BB, BBB).
b) 1 car with exactly 2 girls (GG), then, the other 2 cars are (GBB, BBB) (one of the remaining girls in one of the two remaining cars).
c) 1 car with 2 girls and a boy (GGB), the other 2 cars can have either ->
------> GB and BBB or
------> BB and GBB

(2) First, from the 3 cars, let us select the car that will have only 2 children in it. 3C1 = 3 ways to fix this car.

(3) Now, from the remaining 2 cars (which will have 3 children), we need to fix the car where we will ->

a) either have 3 girls (GGG) or
b) have the remaining girl after assigning 2 girls to the car having only 2 girls (GG GBB BBB) or
c) have 2 girls and a boy (GGB) 

For all the above purposes, we will need to choose one car from the 2 remaining cars. 2C1 = 2 ways to fix this other car. 

Now that we have estimated the number of ways the cars can be selected for our purposes (3 ways to select the 2-seater, and 2 ways to select the main 3-seater = 3 x 2 = 6 ways total), we can focus on estimating the ways in which girls and boys can be selected to fill the cars, based on the scenarios above.


(4) GGG;BB;BBB -> (3C3) x (5C2) x (3C3) = 1 x 10 x 1 = 10 ways.

(5) GG;GBB;BBB -> (3C2) x (1C1 x 5C2) x (3C3) = 3 x 1 x 10 x 1 = 30 ways.

(6) GGB;GB;BBB -> (3C2 x 5C1) x (1C1 x 4C1) x (3C3) = 15 x 4 x 1 = 60 ways.

(7) GGB;BB;GBB -> (3C2 x 5C1) x (4C2) x (1C1 x 2C2) = 15 x 6 x 1 = 90 ways.

(8) Total number of ways of accomodating either 2 or 3 girls in one of the cars = Number of ways of selecting the cars x Number of ways of selecting the children to place in the cars

Total number of ways of accomodating either 2 or 3 girls in one of the cars = (2) x (3) x [ (4) + (5) + (6) + (7) ]

= 3 x 2 x [ 10 + 30 + 60 + 90 ] = 6 x [ 190 ] = 1140. Choice B.

 
___
Harsha
Enthu about all things GMAT | Exploring the GMAT space | My website: gmatanchor.com
 

Total scenarios = 3*8C2*6C3 = 3*28*20 = 1680
Number of scenarios that 1 girl in 1 car = 3!*3*5*4C2 = 6*3*5*6 = 540

=> 1680 - 540 = 1140 -> B