JEE Main 2025 Jan 29 Shift 2, Physics Q20: The Line Spectra of the Hydrogen Atom

From an excited state with principal quantum number $n$, the maximum number of distinct downward transitions (and thus distinct emission lines) is $\left(\genfrac{}{}{0ex}{}{n}{2}\right)=\frac{n(n-1)}{2}$.

For $n=4$: $\frac{4\cdot 3}{2}=6$ lines.

Explicitly: $4\to 3,4\to 2,4\to 1,3\to 2,3\to 1,2\to 1$.