Marbles in a Certain Order

There are $\binom {10}{4} = 210$ ways to choose the spots for the red, blue, green, and orange marble.

For the remaining $6$ spots, there are $6! = 720$ ways to arrange these marbles.

Since these are independent of each other, we can simply multiply to obtain a result of $210 \cdot 720 = \boxed{151200}$ .

There are 10! rearrangements without restrictions. Each of those would contain the desired sequence with the other marbles arranged somehow. Keep those fix and start permuting the red, blue, green and orange ones. These permutations would have all but 1 impossible cases. Then there is a 4!-to-1 function from the set of all 10! permutations into the set of all permutations where we must have $\textcolor{#D61F06}{R}, \textcolor{#3D99F6}{B}, \textcolor{#20A900}{G}, \textcolor{#EC7300}{O}$ in that order. Using the division principle, we have $10!/4!=151200$ desired arrangements.