What set of $10$-digit numbers with $k$ prime factors has largest cardinality? 
*

*Suppose $s_{1}$ are the numbers with 10-digits that have $1$ prime factor.

*Suppose $s_{2}$ are the numbers with 10-digits that have $2$ prime factors.

*Suppose $s_{n}$ are the numbers with 10-digits that have $n$ prime factors.


Leading $0$s are allowed. Which of $s$ would have the smallest expected value of tries to guess the right $10$ digit number?
 A: There are about
$$\frac{10^{10}(\log\log 10^{10})^{k-1}}{(k - 1)!\log 10^{10}}$$
10-digit numbers with k prime factors, allowing leading zeros, or
$$\frac{1}{(k - 1)!}\left( \frac{10^{10}(\log\log 10^{10})^{k-1}}{\log 10^{10}}-\frac{10^9(\log\log 10^9)^{k-1}}{\log 10^9} \right)$$
without leading zeros.
Based on this you'd expect those with 3 or 4 prime factors to be most likely.  The ones with two and five are pretty far behind, with the others not even close.
Counts (with leading 0s):
1 455052511 (predicted: 434294482)
2 1493776443 (predicted: 1362215689)
3 2227121996 (predicted: 2136374810)
4 2139236881 (predicted: 2233663566)
5 1570678136 (predicted: 1751537079)
6 977694273 (predicted: 1098780384)

So those with three prime factors win.  Note: the exact counts assume that you want to count repeated prime factors more than once (the predictions are basically the same either way).  If you look at only the number of distinct prime factors the totals vary but 3 remains in first place.
A: Wolfram Alpha gives the number of up to 10 digit primes as 455,052,511.  When you ask for 2 prime factors, do repeated factors count?  So how many prime factors does $2^{31}*3=6442450944$ have?
Added:  OEIS gives the number of up to 10 digit semiprimes as 1493776443  I don't find a result for other numbers of factors easily.
Edited to allow numbers with leading zeros in the 10 digits
