What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to ground state?
Since comments caused certain level of confusion, I guess I'll try to provide a further illustration.
You should consider all possibilities for an electron "jumping" down the excited energy state $n$ to the ground state $n = 1$.
Electron doesn't get stuck forever on any of the levels with $n &gt; 1$.

Besides that, spectra is not a characteristic of a single excited atom, but an ensemble of many and many excited hydrogen atoms.
In some atoms electrons jump directly from $n = 6$ to $n = 1$, whereas in some others electrons undergo a cascade of quantized steps of energy loss, say, $6 → 5 → 1$ or $6 → 4 → 2 → 1$.
The goal is to achieve the low energy state, but there is a finite number of ways $N$ of doing this.

I put together a rough drawing in Inkscape to illustrate all possible transitions*:



I suppose it's clear now that each energy level $E_i$ is responsible for $n_i - 1$ transitions (try counting the colored dots).
To determine $N$, you need to sum the states, as Soumik Das rightfully commented:

$$N = \sum_{i = 1}^{n}(n_i - 1) = n - 1 + n - 2 + \ldots + 1 + 0 = \frac{n(n-1)}{2}$$

For $n = 6$:

$$N = \frac{6(6-1)}{2} = 15$$

Obviously the same result is obtained by taking the sum directly.



* Not to scale; colors don't correspond to either emission spectra wavelenghts or spectral series and solely used for distinction between electron cascades used for the derivation of the formula for $N$.