solving combination problem     For a certain style of new automobile, the colors
blue, white, black, and green are in equal demand.
Three  successive  orders  are  placed  for  automo-
biles of this style. Find the following probabilities
a  One  blue,  one  white,  and  one  green  are
ordered.
b  Two blues are ordered.
c  At least one black is ordered.
d  Exactly  two  of  the  orders  are  for  the  same
color.

Final answers are there , but i wanna how to solve them by steps :
1) 0.09375
2) 0.140625
3) 0.578125
4) 0.5625
for letter a. i know how to get the total number of outcomes which is simply 4C1 and since u do it 3 times, you cube it. (which turn out to be 64 total outcomes). how do we get / lockdown the success? cant figure out. tnx
edited - for number 1, we get 1/4 , but since we ordering doesnt not matter, we can choose blue, green, white. and since there are 6 possibilities. we multiply it by 6.
for letter b , c , d. i cant still figure it out. 
 A: Note the three orders are successive, there is a first order, then a second, then a third.
I will call an "outcome" the particular sequence of colors for the three orders.
We will work the problem with this in mind.
I will just compute the number of desired outcomes for the different parts of the problem. Of course, to find the probabilities, you just take the number of desired outcomes and divide by the total number of outcomes (assuming outcomes are equally likely).
The total number of outcomes is
$$
\textstyle\bigl({\text{possible colors}\atop\text{for the first order}}\bigr)
\cdot\bigl({\text{possible colors}\atop\text{for the second order}}\bigr)
\cdot\bigl({\text{possible colors}\atop\text{for the third order}}\bigr)=4\cdot4\cdot4
$$

For part a), you need to count the number of outcomes where you know there is  one blue, one white, and one green car. The three colors are determined; so, we just need to count the possible orderings of the colors; for example, the first was green, the second white and the third blue. The total number of outcomes here  is
$$
\textstyle\bigl({\text{choices for}\atop\text{first color}}\bigr)
\cdot\bigl({\text{choices for}\atop\text{  second color}}\bigr)
\cdot\bigl({\text{choices for}\atop\text{  third color}}\bigr)=3\cdot2\cdot1
$$

For part b): two blues are ordered. Here you need to be careful. Is it exactly two blues ordered? If so, then there are three choices for the other color. For each of these there are three different ways to arrange the three individual car orders. For example, if the other color was  white the possible orders are $WBB$, $BWB$, and $BBW$. 
So, in this case, there are $3\cdot3=9$ different outcomes.

For part c), it would be easier to find the number outcomes for which it is not the case that there is at least one black, and subtract this from the total number outcomes:
$$
\textstyle\bigl({\text{number with at }\atop\text{least one black }}\bigr)
=\textstyle\bigl({\text{total number of }\atop\text{possible orders}}\bigr)-
\textstyle\bigl({\text{number with  }\atop\text{no black }}\bigr)
=4\cdot4\cdot4 - 3\cdot3\cdot3
$$
For part d): Here we have exactly two of the orders are for the same color.  We did something similar in part b). There we found that the number of outcomes  where exactly 2 of the orders were for blue was 9. In fact, the  number of outcomes where exactly 2 of the orders were the same particular color is 9. 
Then
$$\eqalign{
\textstyle\bigl({\text{number with  exactly  }\atop\text{two of the same color }}\bigr)
&=
\textstyle\bigl({\text{num. with  exact.  }\atop\text{  two  white }}\bigr)+
\textstyle\bigl({\text{num. with  exact.  }\atop\text{  two  green }}\bigr)+
\textstyle\bigl({\text{num. with  exact.  }\atop\text{  two  blue }}\bigr)+
\textstyle\bigl({\text{num. with  exact.  }\atop\text{  two  black }}\bigr)\cr &
=9+9+9+9.}
$$
(note in the above, we added, since we broke the event "exactly two of the same color" into mutually exclusive events).
