If you solve the wave equation, you will find that the general solution can be written as a linear superposition of the normal modes or harmonics $\sin(k_n x)$ as:
$$y(x,t) = \sum_{n} c_n \sin(k_n x) \cos(\omega_n t),$$
This solution is for the case when the string has some initial displacement (i.e., it is plucked) at $t=0$, and $k_n = n \pi/L$.
Now, the $c_n$s are the "weights" of the different harmonics in the final solution. However, as you can see, they are independent of $t$, and so a different set of $c_n$s would represent a different solution on the string. So how would we normally find the different $c_n$s? Well, we'd use the initial conditions ("How the string was plucked") to do this, since if the string was plucked with some profile $y_0(x)$ at $t=0$, then we would require that: $$y_0(x) = \sum_n c_n \sin\left( \frac{n\pi x}{L}\right) \quad \quad \implies \quad \quad c_n = \frac{1}{L}\int_0^L y_0(x) \sin\left( \frac{n\pi x}{L}\right) \text{d}x.$$
Thus in general, different forms of $y_0(x)$ give different $c_n$s, meaning a different set of excited harmonics. So it's not just where the string was plucked that would excite different harmonics, but also how it was plucked (the "shape" of the initial displacement).
If you want a simple "intuitive" explanation, imagine that you could pluck a string in such a way that you created a node at $x = L/2$. (This can be done quite easily on most guitars.) Since the final solution must be a linear combination of the harmonics, it cannot contain any that do not have a node at $x=L/2$, since they would all have non-zero amplitudes there. As a result, all harmonics with an even number of nodes are eliminated from the solution (i.e., $c_0 = c_2 = c_4 =\dots= 0.$ A similar argument can be made for a string plucked at $x=L/3$, and so on.
