For large numbers, there is a density function which is the probability that a number "near" $n$ is prime. This is $\large \frac{1}{\ln(n)}$ This allows to estimate the number of primes in the interval, lets say, $[10^{49},10^{49}+10^{10}]$ without calculating them.
Do we have such a density function also for the numbers with the property $\ \omega(n)=k\ $ for a given $k$ , where $\ \omega(n)\ $ denotes the number of distinct prime factors of $n$ ? For example, can we estimate how many numbers in the interval, lets say, $\ [10^{49},10^{49}+10^{10}]\ $, have exactly $10$ distinct prime factors ?
$$2\sum_{n \le x, \omega(n) = 2} 1 \sim \sum_{p \le x}\pi(x/p) \sim \sum_{2 \le n \le x} \frac{1}{\log n}\frac{x/n}{\log (x/n)} \sim \frac{x}{\log x} \log\log x$$ and $$\sum_{n \le x, \omega(n) = k} 1 \sim \sum_{p \le x} \sum_{n \le x/p, \omega(n) = k-1} 1 \sim \frac{x}{\log x} \frac{1}{k!} (\log\log x)^k$$
In other words $\sum_{n, \omega(n) = k} n^{-s} \approx P(s)^k \approx (\log \zeta(s))^k$ and $\zeta(s) = \sum_{k=0}^\infty \frac{1}{k!}(\log \zeta(s))^k$
About $[x,x+10^{10}]$ this is a short interval. Those asymptotics are useful to estimate on intervals $[x,(1+\delta) x]$, see the article about the PNT.