$$ \binom{n}{1}\cdot\binom{n-1}{1} = n \cdot (n-1), \qquad\text{whereas}\qquad \binom{n}{2} = \frac{n(n-1)}{2}. $$
It's quite obvious that LHS $\ne$ RHS.
But intuitively we know, as per the definition of $\binom{n}{r}$,
$\binom{n}{1}\cdot\binom{n-1}{1}$ is the equivalent of selection $1$ out of '$n$' objects, followed by selecting $1$ out of '$n-1$' objects (since we have already selected $1$ object, the total number of objects is now '$n-1$').
And, $\binom{n}{2}$ is the equivalent of selecting $2$ out of 'n' objects. Since we're selecting $2$ different objects, it should be the same as selecting $1$ out of '$n$' objects, then another $1$ out of '$n-1$' objects.
I'm sure there is some sound logic for the difference between the $2$ expressions, but it isn't occurring to me. I would be grateful if someone could enlighten me on the reason for the difference betweeen the two.
Here is my question: Why is $\binom{n}{1}\binom{n-1}{1} \neq \binom{n}{2}$?