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Fyodor was throwing paint buckets of equal size in pigeon holes of varying size after a bad day at work. He has one red bucket, two yellow buckets, and three blue buckets. If there are three pigeon holes, the first of which can hold one bucket, the second can hold two buckets, and the third can hold three buckets, how many ways can Fyodor distribute the six buckets among the holes? Note that order of buckets in each hole matters (height order of different colored buckets), and that same color buckets are not distinguishable

So I used casework and after a lot of computation, I get 60. I was wondering if there is a faster way to do this... Interesting enough 60 is $5!/2$.

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    Recognize that any arrangement of buckets in holes can be uniquely described via an arrangement of the word $RYYBBB$ of which there are $\binom{6}{1,2,3}=\frac{6!}{1!2!3!}$ such arrangements. – 2017-02-15

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