Atelevision game has 6 shows doors, of which the contests must pick 2. behind two of the doors are expensive cars, and behind the other 4 doors are consolation prizes. find the probability that the contestant wins exactly 1 car? no car? or atleast one car? what if they had to pick 3 doors? and all of the other information wast the same

The answer to this question:
One car probability 82/120
No car probability = 24/120
At least one car probability= 96/120

I will focus answering the 3 doors probability since the 2nd door problem is solved in the previous problem. (link)

No car condition
1. 1st door consolation, 2nd door consolation=, 3rd door consolation= 4/6 * 3/5 * 2/4= 24/120
This was also can be found by: (4!/1!)/ (6!/3!) = 24/120

(At least one car probability)  is the opposite of (no car probability) In this case, the easier way is 
100% - (no car probability) = 120/120 - 24/120= 96/120

One car probability is (At least one car probability) - (2 car probability). It will be easier to count the 2 car probability and subtract the (At least one car probability) 
Two car condition:
1. 1st door car, 2nd door car, 3rd door consolation = 2/6 * 1/5 * 4/4 =8/ 120
2.1st door car, 2nd door consolation, 3rd door car =2/6 * 4/5 * 1/4 = 8/120
3. 1st door consolation, 2nd door car, 3rd door car= 4/6 * 2/5 * 1/4= 8/120
The total probability will be 8/120+ 8/120 + /120= 24/120
This was also can be found by: (2!) (4!/2!)/ (6!/3!) = 24/120

One car probability =  (At least one car probability) - (2 car probability)= 96/120-24/120= 82/120

Yup correct because it right