# In an FSA, can you have more than one transition for the same symbol from a single state?

Consider the sets of strings on $$\{0,1\}$$ where the 4th symbol from the right end is different from the leftmost symbol. Construct an accepting FSA.

The answer it provides is below. However, I don't understand how you can have one transition for 0,1 and another for 1 from the same state. What does this mean?

The FA in the image is an NFA as clearly a DFA cannot have more than one move for a symbol from a particular state. Also, it doesn't have any moves for state $$q_5$$ nor for $$q_{10}$$.