There are 6 cars lined up next to each other, including: 1 green car, 2 yellow cars and 3 red cars. Calculate the probability that 2 vehicles of the same color are not stacked side by side

Step-by-step explanation:

Total number of ways to arrange the cars

= 6! / (2! * 3!) = 60.

We represent the cars as G, Y and R.

We first arrange the G and Y cars in a row. There are 2 possible cases to investigate.

Case 1: YGY arrangement

The yellow cars are already not stacked side by side.

We have _Y_G_Y_, where the 4 empty spaces are possible places for the red cars to settle. Number of ways = 4C3 = 4.

Case 2: YYG or GYY arrangement

In order for the yellow cars to not be stacked side by side, a red car must be between them.

We have _YRY_G_ or _G_YRY_, where the 3 empty spaces are possible places for the remaining 2 red cars to settle.

Number of ways = 2 * 3C2 = 6.

Hence the probability of such an event

happening is (4 + 6)/60 = 1/6.

The answer is 1/6.

7. d. 64

8. i. 32

9. c. 125

10. j. none of the above

11. d. 81

12. i. 216

13. e. none of the above

14. g. 240

15. e. none of the above

16. g. 32

17.   e. none of the above

c.

step-by-step explanation:

because first you have to add the 4x, 9x, and the 7x together and you get 20x and then you add the last numbers up and you get 5and then you put the subtraction sign in there and that's how you get the answer.