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 difference. 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.

Final Answer:

a =  90
b =  3

Step-by-step explanation:

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

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

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