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2 Answers

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nos of ways fromA to C via B is 4*3=12

Similarly no of ways from C to A via B =3*4=12.

So for round trip nos of ways =12*12=144

Hence the option C is correct.
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Number of ways to choose path from $A$ to $B$ = 4C1 

Number of ways to choose path from $B$ to $C$= 3C1

Now during roundtrip

Number of ways to choose path from $C$ to $B$ = 3C1

Number of ways to choose path from $B$ to $A$= 4C1

Total possibe combinations= $4C1 * 3C1 * 3C1 * 4C1 $= $4*3*3*4= 144$

Hence Option C) is correct answer

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