How many sets of two factors of 360 are coprime to each other? My attempt:
$360=2^3\cdot3^2\cdot5^1$
Number of sets of two factor coprime sets for $2^3$ and $3^2$ only $=12+6=18$
With that if we add the effect of $ 5^1$, number of sets $=18+2\cdot 18-1=53$.
Is this ok? 
The answer given is $56$.
 A: First suppose that $5$ divides neither coprime factor. Then we have two cases- 
Case 1: One factor must be $2$ to some positive power while the other factor must be $3$ to some positive power. Hence there are $3\cdot2=6$ of these (as there are $3$ possible positive powers of $2$ and there are $2$ possible positive powers of $3$). 
Case 2: One factor is $1$ while the other factor is a divisor of $2^33^2$. Hence there are $(3+1)(2+1)=12$ of these.
Now, note that all of these pairs consist of different numbers except the pair $\{1,1\}$. For each of the pairs with two distinct elements, we can multiply either element by $5$ to get two new coprime pairs of factors (that is the solution $\{a,b\}$ gives rise to the solutions $\{5a,b\}$ and $\{a,5b\}$). Finally, the pair $\{1,1\}$ under this process only gives us one new pair: $\{1,5\}$.
In conclusion, we have
$$3(6+12-1)+2=53$$
coprime pairs.
A: I'll assume you mean sets with exactly two elements, so the two factors $1$ and $1$ (represented by the singleton $\{1\}$) won't count. 

Case 1: One of the factors is $1$. 
$\gcd(1,n)=1$ for all $n$, so this is just the number of factors of $360$, minus $1$. The number of factors of $360=2^3\cdot3^2\cdot5$ is simply the product of one plus the exponents on the primes (think of the number of choices for the multiplicity of each prime factor), which is $4 \cdot 3 \cdot 2 = 24$. 
This gives the number of sets in this case as $24-1=23$. 
Case 2: Neither of the factors is $1$, but both are odd. 
This means we choose a power of $3$ and a power of $5$. There are $2$ choices for a power of $3$ and $1$ for a power of $5$ (we don't count $3^0=5^0=1$ in this case). There are $2$ such sets.
Case 3: One of the factors is even, and the other is not $1$. 


*

*If the even number has no odd prime factors, then there are $3$ choices for it: $2, 4,$ and $8$. Given this, there are $3 \cdot 2 -1$ choices for the other factor. This gives $15$ for this sub-case.

*If the even number is divisible by three, then there are $6$ choices for it. Given this, there is only $1$ choice for the other factor: $5$. There are $6$ possibilities in this sub-case.

*If the even number is divisible by five, then there are $3$ choices for it. Given this, there are $2$ choices for the other factor: $3$ or $9$. This gives $6$ possibilities for this sub-case.
There are a total of $15+6+6=27$ possibilites in this case.

Our total is $23+2+27=\boxed{52}$. This doesn't agree with your given answer.
A: for a given $n$ let $T(n)$ represent the total number of unordered pairs of distinct factors of $n$, and let $A(n)$ represent the cardinality of the subset of these in which the two factors are mutually prime. 
since each pair of factors has a greatest common divisor which is a factor of $n$ we have
$$
T(n) = \sum_{d|n} A(\frac{n}{d}) = \sum_{d|n} A(d)
$$
and by the Mobius inversion theorem we have:
$$
A(n) =  \sum_{d|n}\mu(d) T(\frac{n}{d}) \tag{1}
$$
if a number has $k$ factors, then it has $\frac12 k(k-1)$ unordered pairs of distinct factors. substituting for $T(n)$ we find that:
$$
A(360)=\frac12(24\cdot 23-18\cdot 17 - 16 \cdot 15 -12 \cdot 11 + 12 \cdot 11 +9 \cdot 8 +8\cdot 7 -6 \cdot 5) = 52
$$
adding the pair $(1,1)$ gives a final tally of 53, in agreement with other estimates here
