# Restricted arrangements of six 0's, five 1's and four 2's

The task is to find the # of arrangements of six 0's, five 1's and four 2's with the restriction that the first 0 precedes the second 1.

I have worked it out by finding total # of arrangements (630630), and subtracting invalid arrangements

(i) firstly arranging 11 followed by the remaining 1's & 0's

(ii) fitting the 2's into the "interstices" in various patterns

I get a (hopefully correct) answer of 515970, but the process seems crude, and the question is:

Is there some slicker way of solving it ?

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Then the final answer is $${15\choose11}\left({11\choose5}-{9\choose3}\right)$$