How many ways can 6 cars ( 3 pink, 2 orange and 1 yellow) be parked in 6 parking slots in a row? a. If the pink cars must be park together? - my answer is 4!3! or 144
b. If the orange cars must not be parked together?
c. If you can't park the yellow on either end?
d. If a pink car must be on both ends?
b,c,d is quite confusing tho
Cars of the same color are distinguishable.
 A: Treating the cars as distinct, here is another approach:
(a) you have solved correctly.
(b) permute non-orange cars in 4! ways, and place the orange cars in the gaps in between including end gaps:  _C_C_C_C_ in $5\cdot4$ ways to get $20\cdot4!$ ways
(c) again, permute non-yellow cars in 5! ways and the yellow in the 4 in-between gaps C_C_C_C_C, to get 4*5! ways
(d) last one given without explanation: $_3P_2 \times 4!$ ways
A: Since cars of the same color are distinguishable, your answer for (a) is correct.
To determine the number of ways the cars can be parked if the two orange cars are not adjacent, we subtract the number of arrangements in which they are adjacent from the total number of arrangements.  There are $6!$ ways to park six distinct cars in a row.  If the orange cars were together, there would be five objects to arrange (the block of two orange cars and the other four cars), which can be done in $5!$ ways.  Since the two orange cars can be arranged in $2!$ ways, there are $5!2!$ arrangements in which the two orange cars are adjacent.  Hence, there are $6! - 5!2!$ arrangements in which they are not adjacent.  
If the yellow car is parked at one of the two ends, there are $5!$ ways of parking the other cars.  Hence, there are $6! - 2 \cdot 5!$ arrangements in which a yellow car is not parked at either end.
There are $3$ ways to pick which pink car will be at the left end and $2$ ways to pick which of the remaining pink cars will be at the right end.  For each such choice, there are $4!$ ways of parking the other four cards.  Hence, there are $3 \cdot 2 \cdot 4!$ arrangements in which a pink car is parked at each end of the row.
A: a. If the pink cars must be park together? - my answer is $4!3!$ or $144$. Treat the $3$ pink cars as a single object.  You have $4$ objects, the pink cars as a whole the two orange cars, the $1$ yellow so there are $4!$ ways to order them.  There are then $3!$ ways to order the pink cars, so, as you say, $3!4!$ ways to do this.
b. If the orange cars must not be parked together? There are, of course, $6!= 720$ ways to park $6$ cars with no conditions.  There are, as above, $5!2!= 240$ ways to park the cars with the $2$ orange cars parked together so there are $6!- 5!2!= 480$ ways to park the cars with the orange cars not together. 
c. If you can't park the yellow on either end? There are $5$ non-yellow cars so $5$ choices for the car on the left.  There are $4$ non-yellow cars left so $4$ choices for the car on the right.  There are now $4$ cars left, including the yellow car which can be parked in any of the $4$ places left so $4!$ ways to park those.  There are $5(4)(4!)= 480$ ways to do this.
d. If a pink car must be on both ends? With a pink car on each end, there are $4$ cars left which can be parked in any of the remaining $4$ places- there are $4!= 24$ ways to do that.
