A silly approach would be to just count them. You can include at most two elevens, when we do this the product of the remaining numbers in the prime factorization must be at most four, so how many do we get in this case? If a product contains exactly one 11, the product of the rest of the prime factorization must be at most 45 so how many do you get in this case? etc.

The answer is $32$?

I got $31$... maybe I just can't do addition?

$n=2^a7^b11^c$ and $c<3$ Case 1: $c=2, 2^a7^b\leq 4 \to (a,b,c)=(0,0,2),(1,0,2)(2,0,2)$ Case 2: $c=1,2^a 7^b\leq 45 \to b \leq 1 \to (a,b,c)=(0,1,1)(1,1,1),(2,1,1),(a,0,1)$, where $0\leq a \leq 5$ Case 3: $c=0 \to b \leq 3$ Case 3.1: $b=3 \to a=0$ Case 3.2: $b=2 \to 0 \leq a \leq 3$ Case 3.3 $b=1 \to 0 \leq a \leq 6$ Case 3.4: $b=0 \to 0 \leq a \leq 8$ So total we have $3+3+6+1+4+7+9=33$