The $2^{nd}$ last nodes before $I$ are $F$ , $H$ and $E$.
Through F ,1 way $F\rightarrow I$
Through H, 2 ways $H\rightarrow I$ or $H\rightarrow F\rightarrow I$.
Through E, 5 ways.
- $ E\rightarrow C\rightarrow F\rightarrow I$ and from F its 1 way.
- $ E\rightarrow H $ and from H its 2 ways.
- $ E\rightarrow F $ and from F its 1 way.
- $E\rightarrow I$ 1 way
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Now lets fix the $1^{st}$ and $2^{nd}$ node from the beginning and see all the cases possible.
- A $\rightarrow$ B ..... I ( keeping A $\rightarrow $ B fixed )
- A $\rightarrow $ B $\rightarrow $ F ....I => 1 way.
- A $\rightarrow $ B $\rightarrow $ E .... I => 5 ways.
- A $\rightarrow $ B $\rightarrow $ C .... I => 1 way.
$\therefore$ 7 ways keeping A $\rightarrow $ B fixed.
- A $\rightarrow$ E ..... I ( keeping A $\rightarrow $ E fixed. )
$\therefore$ 5 ways keeping A $\rightarrow $ E fixed.
- A $\rightarrow$ D ..... I ( keeping A $\rightarrow $ D fixed. )
- A $\rightarrow $ D $\rightarrow $ E ....I => 5 ways.
- A $\rightarrow $ D $\rightarrow $ H ....I => 2 ways.
- A $\rightarrow $ D $\rightarrow $ B ....I => 7 ways.
- A $\rightarrow $ D $\rightarrow $ G ....I
- G $\rightarrow $ E .... I => 5 ways.
- G $\rightarrow $ H....I => 2 ways.
$\therefore$ 5+2+7+5+2 = 21 ways keeping A $\rightarrow $ D fixed.
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$\therefore$ Total number of paths possible from A $\rightarrow $.....I $= 7+5+21 = 33 $