A FSA has the following transition function:

$\{0->1($a$),0->2($b$),0->3($c$),0->6(</>),$

$1->2($b$),1->3($c$),1->6(</>),$

$2->1($a$),2->3($c$),2->6(</>),$

$3->3($a$),3->3($b$),3->3($c$),3->5(</>),$

$4->4($a$),4->4($b$),4->4($c$),4->6(</>)\}$

Knowing that the alphabet is $\{$ a$,$ b$,$ c $\}($plus the end of the word $</>$ symbol$),$ the initial state has no incoming arc, and the final states have no exiting arcs, how many states have this FSA?

A FSA has the following transition function:

$\{0->1($a$),0->2($b$),0->3($c$),0->6(</>),$

$1->2($b$),1->3($c$),1->6(</>),$

$2->1($a$),2->3($c$),2->6(</>),$

$3->3($a$),3->3($b$),3->3($c$),3->5(</>),$

$4->4($a$),4->4($b$),4->4($c$),4->6(</>)\}$

Knowing that the alphabet is $\{$ a$,$ b$,$ c $\}($plus the end of the word $</>$ symbol$),$ the initial state has no incoming arc, and the final states have no exiting arcs, how many states have this FSA?

$7$ total states, being $2$ initial states, and $3$ final states.

$7$ total states, being $1$ initial state, and $2$ final states.

$6$ total states, being $1$ initial state, and $3$ final states.

$6$ total states, being $2$ initial states, and $2$ final states.