Seven different models cars, from which there are three blue and four red, are to be parked in a row. Find how many different arrangements there are if no two cars of the same color are to be parked next to each other.

A blue car cannot be on either end.

So we have to start and end with a red car.
They must be parked this way

RBRBRBR

number of way to do this is
R-B-R-B-R-B-R
4x3x3x2x2x1x1
= 144

To find the number of different arrangements, we need to consider the positions of the cars.

Let's start by placing the blue cars. We have 3 blue cars to be placed in 3 positions. Since no two cars of the same color can be parked next to each other, we can place the blue cars in the following ways:

1. B _ B _ B _
2. _ B _ B _ B
3. B _ B _ _ B
4. _ B _ B _ _

For each arrangement, there are 3! = 6 ways to arrange the blue cars within the allotted positions.

Now let's place the red cars. We have 4 red cars to be placed in the remaining empty positions (marked with "_"). Since no two cars of the same color can be parked next to each other, we can place the red cars in the following ways:

1. R _ R _ R _ R
2. _ R _ R _ R _

For each arrangement, there are 4! = 24 ways to arrange the red cars within the allotted positions.

Multiplying the number of arrangements for the blue cars (6) with the number of arrangements for the red cars (24), we get:

6 * 24 = 144

Therefore, there are 144 different arrangements if no two cars of the same color are to be parked next to each other.

Let's solve this problem step by step:

Step 1: Determine the positions of the blue cars.
Since there are 3 blue cars, we can choose 3 positions for the blue cars out of the 7 total positions. We can do this using combinations: C(7, 3) = 35.

Step 2: Arrange the blue cars.
Now let's arrange the 3 blue cars in the 3 chosen positions. Since all 3 blue cars are different models, we can arrange them in 3! = 6 ways.

Step 3: Arrange the red cars.
Since there are 4 red cars, we need to place them in the remaining 4 empty positions. We can do this in P(4, 4) = 4! = 24 ways.

Step 4: Multiply the results.
Finally, we need to multiply the results from steps 2 and 3 to obtain the total number of arrangements. Multiplying 6 by 24 gives us 144.

Therefore, there are 144 different arrangements if no two cars of the same color are to be parked next to each other.