Find the total number of factors (excluding $1$) of $2160$.
According to me it's just asking about total number of factors that is $4+3+1=8$ ($2^4 \cdot 3^3 \cdot 5^1$). I don't see any statement asking about number of ways of selecting its factors.
But solution of this question is given by taking number of ways of selecting some or all out of four $2$s,three $3$s and one $5$. Number of such ways is $(4+1)(3+1)(1+1)-1=39$.
Where am I wrong ?
On
Let's not work with such a big number, let's work with $6 = 2^1 \times 3^1$.
By your argument, you are claiming that there are only $1+1=2$ factors. Perhaps you are just counting $2$ and $3$ and not counting that it is possible to select $6$ as a potential factor.
It seems that you are computing factors of the form of $p^r$ where $r \ge 1$ where $p$ is a prime which is not what the question is asking for.
In general if the prime factorization is $\prod_{i=1}^n p_i ^{m_i}$, then the number of factor is $\prod_{i=1}^n (m_i+1)$ . If you are not counting $1$ as a factor, then we subtract by $1$. The rationale of the formula is for each prime, we can consider how many of copies of a particular prime should be chosen and it is possible that we choose none.
On
Divide the prime factors into three groups: $$2^\color{red}4=1\cdot 2\cdot 2\cdot 2\cdot 2; \ \ \ \ 3^\color{green}3=1\cdot 3\cdot 3\cdot 3; \ \ \ 5^\color{blue}1=1\cdot 5.$$ To form a factor you choose a number from each group. Here are the possible numbers in the groups: $$\{1,2,2^2,2^3,2^4\}; \ \ \{1,3,3^2,3^3\}; \ \ \{1,5\}.$$ Note that choosing $1$ basically means you do not choose the prime number (in this case $2,3,5$).
So by the product rule in combinatorics and the fundamental theorem of arithmetic: $$\begin{align}{5\choose 1}\cdot {4\choose 1}\cdot {2\choose 1}&=5\cdot 4\cdot 2=40 \ \ \text{OR}\\ {\color{red}4+1\choose 1}\cdot {\color{green}3+1\choose 1}\cdot {\color{blue}1+1\choose 1}&=5\cdot 4\cdot 2=40.\end{align}$$
Hint:
$$12=2^2\cdot3$$
The possible factors are
$$1,2,4,3,6,12$$ respectively $$2^0\cdot3^0,2^1\cdot3^0,2^2\cdot3^0,\\2^0\cdot3^1,2^1\cdot3^1,2^2\cdot3^1.$$
As you see, every prime factor is taken will all multiplicities, and all combinations ($(2+1)(1+1)$ of them) are possible.