Arranging 7 rows of 3 motercycles of 5 different types and 4 different colours A playground equipment manufacturer makes a biker formation comprised of $21$ wooden
motorcycles that are fixed in place, three abreast, to form $7$ rows of $3$ motorcycles each.
She has $5$ different types of wooden motorcycle that can be installed. All motorcycles
face forwards.
Each motorcycle is painted with one of $4$ different colours. 
How many
different biker formations can be made if:
(a)there are no restrictions?
(b) at least one motorcycle of each type (and some colour) must be installed?
(c)the only restriction is that no two motorcycles in the same row can
have the same colour?
Here is what I attempted (not sure if its right):
a):
$5^3$ for each row (5 selections of bike, 3 bikes each row) so :
$7 * 5^3$ bikes.
There are 4 colours available, so $7*4^3$ colouring. In total then : 
$$7* (5^3 + 4^3)$$
b):
One bike of each model must be installed:
21-5 = 16 bikes left to install:
Don't know how to go on from here?
c):
so first get all the bikes installed:
$7*5^3$ then to colour it $7*(4*3*2)$
Am I doing part a and c right? and what should I do for part b
 A: Since a motorcycle can be of any of $5$ types and $4$ colors, there are altogether $5\cdot4=20$ variants of a single motorcycle. If there are no restrictions, each of the $21$ motorcycles in the formation can be any of the $20$ variants, so filling out the formation involves making $21$ $20$-way choices. This can be done in $20^{21}$ different ways, which is therefore the answer to (a).
We can analyze (c) as follows. There are $5^3$ possible sequences of $3$ types, so if we ignore color, there are $5^3$ possible rows. There are $4\cdot3\cdot2=24$ ways to assign $3$ different colors to the motorcycles in a row of $3$, so when we take colors into account, there are $24\cdot 5^3$ different possibilities for each row. Each row must be one of these $24\cdot 5^3$ possibilities, and there are $7$ rows, so there are $(24\cdot5^3)^7=24^7\cdot 5^{21}$ formations in which no two motorcycles in the same row have the same color.
Probably the easiest way to get the answer to (b) is to compute the number of formations that violate the constraint and subtract that from the answer to (a). The formations that violate the constraint are those that use at most $4$ of the types. If the types are $T_1,T_2,T_3,T_4$, and $T_5$, the analysis in (a) shows that there are $16^{21}$ formations that use only types $T_1,T_2,T_3$, and $T_4$. Similarly, there are $16^{21}$ that use only types $T_1,T_3,T_4$, and $T_5$, and the same number for each of the other $5$ sets of $4$ types. That’s a total of $5\cdot16^{21}$ unacceptable formations, so a first approximation to the desired answer is $20^{21}-5\cdot16{21}$.
Unfortunately, we’ve subtracted too much: any formation that uses only types $T_1,T_2$, and $T_3$ was counted once in the $20^{21}$ term and subtracted twice in the $5\cdot16^{21}$ term, once for using only types $T_1,T_2,T_3$, and $T_4$ and once for using only types $T_1,T_2,T_3$, and $T_5$. Thus, the figure $20^{21}-5\cdot16^{21}$ counts each of these formations a net of $-1$ times, and we must add them back in so that they are counted $0$ times. That means adding back in the $12^{21}$ formations that use only the three types $T_1,T_2$, and $T_3$, and similarly for each other set of three types. There are $\binom53=10$ sets of three types, so our new approximation is $20^{21}-5\cdot16^{21}+10\cdot12^{21}$. Two further corrections are needed to adjust for formations using just two colors and just one color; I’ll leave these to you. What I’m doing here is an inclusion-exclusion argument.
