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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?

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up vote 7 down vote accepted

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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Coprime factors have disjoints sets of prime factors. Hence the number of possible coprime decompositions is the same as half the number of subsets of $P=\{2,3,5,7\}$, which is 8. (Each coprime decomposition appears twice when you list the $16=2^4$ subsets of $P$.)

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Hint $\ $ Coprime factorizations of $\rm\:n\:$ biject with coprime factorizations of the squarefree part of $\rm\:n,\:$ which biject with unordered two-part partitions of the set of prime factors of $\rm\:n\:$.

$$\rm (P_1^{J_1}\cdots P_M^{J_M})\: (Q_1^{K_1} \cdots Q_N^{K_N})\ \leftrightarrow\ (P_1\cdots P_M)\:(Q_1\cdots Q_N)\ \leftrightarrow\ \{P_1,\ldots,P_M\}\cup \{Q_1,\ldots,Q_N\}$$

Note that the result depends heavily on the uniqueness of prime factorizations.

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