2015 AMC 10A Problems/Problem 10

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Solution

Observe that we can't begin a rearrangement with either a or d, leaving bcd and abc, respectively.

Starting with b, there is only one rearrangement: $bdac$. Similarly, there is only one rearrangement when we start with c: $cadb$.

Therefore, our answer must be $\boxed{\textbf{(C) }2}$.