Counting ordered triples whose product is at most $n$ There is a classical question in discrete math such a kind  that

Let $a\times b \times c =24 $ then , how many possible $(a,b,c)$ triples are there where $a,b,c \in Z^+$

The answer is the following $x_1+x_2+x_3 =3$ and $y_1+y_2+y_3 =1$ , then $$\binom{3+3-1}{3}\binom{3+1-1}{1}=30$$
Now , i want to turn this question to this one such that

Let $a\times b \times c < 24 $ then , how many possible $(a,b,c)$ triples are there where $a,b,c \in Z^+$

In first hand , it thought to add another integer $d$ such that $d >1 $ and  $a\times b \times c \times d = 24 $
Now , i must do the same calculation like in triple case and subtract the cases where $d=1$ from the total.For example , $$\binom{4+3-1}{2}\binom{4+1-1}{1}-\binom{3+3-1}{3}\binom{3+1-1}{1}=60-30 =30$$
However , my logic is wrong ,i saw it for small case such as $a \times b \times c =6$.
Do you have any suggestion or solution for my problem except for computer programming ?
B.T.W My question is not only specific for $24$ , i chose it because it is easy to handle , but it could be any other number such as $678$ etc. So , i am looking for the solution which can handle large numbers
 A: Let $f_3(n)$ be the number of positive integer solutions to
$$
a\times b\times c\le n
$$
We can compute $f_3(n)$ in approximately $O(n^{5/6})$ time.
First, let us define $f_2(n)$ to be the number of solutions to $a\times b\le n$, as this will be useful for computing $f_3(n)$. Note that for any pair $(a,b)$ such that $a\times b\le n$, we will have
$$
\text{either $a\le \lfloor n^{1/2}\rfloor $ or $b\le \lfloor n^{1/2}\rfloor $}
$$
Therefore, using the principle of inclusion exclusion,
$$
\begin{align}
f_2(n)
&=\phantom{ + }\#\{\text{pairs where $a\le \lfloor n^{1/2}\rfloor $}\} \\
&\phantom{ = }+\#\{\text{pairs where $b\le \lfloor n^{1/2}\rfloor $}\} \\
&\phantom{ = }-\#\{\text{pairs where $a\le \lfloor n^{1/2}\rfloor $ and $b\le \lfloor n^{1/2}\rfloor $}\}
\end{align}
$$
Conclude as follows:
$$
\begin{align}
\#\{\text{pairs where $a\le \lfloor \sqrt n\rfloor $}\}
&=\sum_{a=1}^{\lfloor n^{1/2}\rfloor} \lfloor n/a\rfloor\\
\#\{\text{pairs where $a\le \lfloor \sqrt n\rfloor $ 
and $b\le \lfloor \sqrt n\rfloor $}\}
&=(\lfloor n^{1/2}\rfloor )^2
\end{align}
$$
Using these formulae, computing $f_2(n)$ takes $O(n^{1/2})$ time, ignoring the cost of arithmetic operations.
Now, let us leverage this to compute $f_3(n)$. Let $E_a$ be the set of triples $(a,b,c)$ such that $a\times b\times c\le n$, and for which $a\le \lfloor n^{1/3}\rfloor$. Define $E_b$ and $E_c$ similalry. Using PIE again,
$$
f_3(n)=|E_a|+|E_b|+|E_c|-|E_a\cap E_b|-|E_a\cap E_c|-|E_b\cap E_c|+|E_a\cap E_b\cap E_c|
\\ \hspace{-5.65cm}=3|E_a|-3|E_a\cap E_b|+|E_a\cap E_b\cap E_c|
$$
Finally, conclude by noting
\begin{align}
|E_a|&=\sum_{a=1}^{\lfloor n^{1/3}\rfloor }f_2(\lfloor n/a\rfloor )
\\
|E_a\cap E_b|&=
\sum_{a=1}^{\lfloor n^{1/3}\rfloor}
\sum_{b=1}^{\lfloor n^{1/3}\rfloor}\lfloor n/(ab)\rfloor
\\
|E_a\cap E_b\cap E_c|&=(\lfloor n^{1/3}\rfloor )^3
\end{align}
A: The original function is known as $\tau_3(n)$.  The sequence is in the OEIS as A007425.  Your question about the number of ordered triples $(a,b,c)$ with $abc < 24$ is  $\sum_{i=1}^{23} \tau_3(n)$.
Those partial sums are A061201 which includes a nice formula given by Wesley Ivan Hurt using the integer floor function:
$$a(n) = \sum_{k=1}^n \sum_{i=1}^n \left\lfloor\frac{n}{ik}\right\rfloor.$$
The derivation is probably along the lines of Mike Earnest's answer.  That formula is for partial sums including $\tau_3(n)$, so the answer to your question for 24 is $a(23) = 173$.
A: This solution is only intended for the problem specified in the title. And this is not the full solution.
We have $$a×b×c<24$$
Consider the case where $a=b=c$
Triplets here formed will be only $$(1,1,1) \:\:(2,2,2)$$
Now consider the case $a=b\ne c$
Triplets formed here will be
$$(1,1,2)\:\:(1,1,3)\cdots(1,1,23)$$
$$(2,2,1)\:\:(2,2,3)\:\:(2,2,4)\:\:(2,2,5)$$
$$(3,3,1)\:\:(3,3,2)$$
$$(4,4,1)$$
Cases $a=c\ne b$ and $b=c\ne a$ will have same number of cases as above
So till now we got $$2+29×3=89 \:\:\textrm{cases}$$
Now the case left is $a\ne b\ne c$
And I'm working on that...
