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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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up vote 1 down vote accepted

There are 15 slots to be filled. 11 will be filled with zeros and ones. Choosing those 11 slots can be done in 15-choose-11 ways.

Having chosen the 11 slots, how many legit ways are there to fill them? Easier to count the illegit ways, those with ones in the first two slots; then the other 9 slots get three ones and six zeros, so 9-choose-3 ways. So the number of legit ways is 11-choose-5 minus 9-choose-3.

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

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+1: Thanks, that was very neat ! Out of curiosity, though, I wonder whether there'd be another way working with base 3. – true blue anil Mar 7 '12 at 6:05

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