Wagons

We have six wagons, two white, two blue, and two red. We assemble trains from them, wagons of the same color are exactly the same, so if we change only two white wagons on a train, it's still the same train, because I don't know any different. How many different trains can we assemble when:
The 1st train must have all six wagons
The 2nd train can have 1 to 6 wagons.

Correct result:

a =  90
b =  3

Solution:

$a={\displaystyle \frac{(2+2+2)!}{2!\cdot 2!\cdot 2!}}=90$

$$\begin{gathered} b_1=1+1+1=3 \\ {b}_2=2+2+2=6 \\ b=3 \end{gathered}$$

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