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?
 A: 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.
$$\{2^3 , 3^2\cdot 7^2\cdot 11\}$$
$$\{2^3\cdot 3^2, 7^2\cdot 11\}$$
$$\{2^3\cdot 7^2, 3^2\cdot 11\}$$
$$\{2^3\cdot 3^2\cdot 7^2, 11\}$$
$$\{2^3\cdot 11 , 3^2\cdot 7^2\}$$
$$\{2^3\cdot 3^2\cdot 11, 7^2\}$$
$$\{2^3\cdot 7^2\cdot 11, 3^2\}$$
$$\{2^3\cdot 3^2\cdot 7^2\cdot 11, 1\}$$
A: 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 $P$.)
A: 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.
A: its 8 because the number of ways in which a composite no. can be resolved as coprimes is 2^(n-1) where n is the no. of distinct prime factors in the expression.
38808=2^(3)+3^(2)+7^(2)+11, so number of distinct primes is 4, so 2^(4-1)=2^3=8
A: As the OP has mentioned the factorization is of the form $$38808 = 2^3 \times 3^2 \times 7^2 \times 11$$
we  have the following options of selecting distinct prime factors.make four groups of numbers  such that one group contains all 2's,second group contains all 3's and so on.now the question involves selection of one group or two groups .this can be done in 4C1+ 4C2
=4+6=10.But here many are counted twice.for example in 4C2 we have counted twice the cofactors because the two groups selected for one coprime factor gets counted again when we go on for selection of remaining two groups.Hence the actual number of coprime factors are 10-2=8
A: The number of ways in which a composite number N can be resolved into two factors which are relatively co-prime to each other is equal to 2^(n-1) .
Where n is the number of different factors of N.
Ex.38808= 2^3 × 3^2 × 7^2 × 11.
Here, n=4, so, no. of ways= 2^(4-1)= 2^3= 8. 
