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Q.
The number of ways of factoring 91,000 into two factors, $m$ and $n$, such that $m >1, n >1$ and $\operatorname{gcd}(m, n)=1$ is
Permutations and Combinations
Solution:
We have $91,000=2^3 \times 5^3 \times 7 \times 13$ Let $A=\left\{2^3, 5^3, 7,13\right\}$ be the set associated with the prime factorization of 91,000 . For $m, n$ to be relatively prime, each element of $A$ must appear either in the prime factorization of $m$ or in the prime factorization of $n$ but not in both. Moreover, the 2 prime factorizations must be composed exclusively from the elements of $A$. Therefore, the number of relatively prime pairs $m, n$ is equal to the number of ways of partitioning $A$ into 2 unordered non-empty subsets. We can partition $A$ as follows:
$\left\{2^3\right\} \cup\left\{5^3, 7,13\right\},\left\{5^3\right\} \cup\left\{2^3, 7,13\right\} $
$\{7\} \cup\left\{2^3, 5^3, 13\right\},\{13\} \cup\left\{2^3, 5^3, 7\right\} $
and $\left\{2^3, 5^3\right\} \cup\{7,13\},\left\{2^3, 7\right\} \cup\left\{5^3, 13\right\}, $
$\left\{2^3, 13\right\} \cup\left\{5^3, 7\right\}$
Therefore, the required number of ways $=4+3=7$.