<s>[INST] Number of ways in which $38808$ can be expressed as a product of 2 coprime factors? Number of ways in which $38808$ can be expressed as a product of $2$ coprime factors ?
the answer given is $8$ ways, 
what I did was,
$$38808 = 2^3 \times 3^2 \times 7^2 \times 11$$
so the number of ways of expressing $38808$ as product of $2$ co-prime factors should be
$$8 \cdot (9.49.11)$$
$$9 \cdot (8.49.11)$$
$$49 \cdot (9.8.11)$$
$$11 \cdot (9.49.8)$$
hence $4$, but the answer is $8$, am i missing some other co-prime factor pairs?
 [/INST] No, 8 is correct.  Look at the prime factorisation.  Two factors would be coprime if the four exponent-expressions in the prime factorisation were partitioned between the two factors.  There are 8 ways of doing this, as follows.
$$\{2^3 , 3^2\cdot 7^2\cdot 11\}$$
$$\{2^3\cdot 3^2, 7^2\cdot 11\}$$
$$\{2^3\cdot 7^2, 3^2\cdot 11\}$$
$$\{2^3\cdot 3^2\cdot 7^2, 11\}$$
$$\{2^3\cdot 11 , 3^2\cdot 7^2\}$$
$$\{2^3\cdot 3^2\cdot 11, 7^2\}$$
$$\{2^3\cdot 7^2\cdot 11, 3^2\}$$
$$\{2^3\cdot 3^2\cdot 7^2\cdot 11, 1\}$$
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