How many ordered triplets (a, b, c) satisfy a^b^c=64 where a, b, c are all positive integers? (10, 7, 8, 9)

64 = $$2^6$$ = $$4^3$$ = $$8^2$$ = $$64^1$$
Case 1: a = 2 => 6 = $$b^c$$, we get b = 6 , c = 1
Case 2: a = 4 => 3 = $$b^c$$, we get b = 3 , c = 1
Case 3: a = 8 => 2 = $$b^c$$, we get b = 2 , c = 1
Case 4: a = 64 => 1 = $$b^c$$, we get b = 1 , c = 1
The total number of such ordered pair is 4. Hence answer is none of the given options
Let us consider the case if the question is $$(a^b)^c$$ = 64 => $$a^{bc}$$ = 64
Case 1: a = 2 => 6 = $$bc$$, we get (b,c) = {(1,6),(2,3),(3,2),(6,1)}
Case 2: a = 4 => 3 = $$bc$$, we get (b,c) = {(1,3),(3,1)}
Case 3: a = 8 => 2 = $$bc$$, we get (b,c) = {(1,2),(2,1)}
Case 4: a = 64 => 1 = $$bc$$, we get (b,c) = {(1,1)}
Hence total number of ordered pairs is 9.
 

Hi, is that $$a^{b^c}$$ or $$(a^b)^c$$?

first one, not the second one. And answer is 9