Find the number of all possible distinct pairs of two coprime integers such that their product is 180.

Prime factorization of 180:

$180 = 2^2 \times 3^2 \times 5$

For two numbers to be coprime, they must not share any common prime factor.
So, each prime power must go completely to only one of the two numbers.

There are 3 distinct prime factors: $2,;3,;5$.

For each prime factor, we have 2 choices:

• either it goes to the first number, or

• it goes to the second number.

Number of ordered pairs
$= 2^3 = 8$

But the question asks for distinct pairs (order does not matter), so
$(a,b)$ and $(b,a)$ are the same.

Since $a=b$ is impossible (they would not be coprime),

Number of distinct pairs $= \frac{8}{2} = 4$