Number of spectral lines from $n_2$ to $n_1$

Question

When electrons de-excite from higher energy level $\left(n_{2}\right)$ to lower energy level $\left(n_{1}\right)$ in atomic sample, then number of spectral line observed in the spectrum $$ =\frac{\left(\mathrm{n}_{2}-\mathrm{n}_{1}\right)\left(\mathrm{n}_{2}-\mathrm{n}_{1}+1\right)}{2} $$

According to Quora, this is derived from permatuation & combination. Can you please help me the derivation from permatuation & combination

Answer 1

Let $k=n_2-n_1$. the number of states between (and including) energy levels $n_1$ and $n_2$ is $k+1$. thus the number of spectral lines is the number of ways of choosing 2 states from the $k+1$ states, which is $${k+1\choose 2}=\frac{(n_2-n_1+1)(n_2-n_1)}{2}$$

Answer 2

When electron excited to n state. It can be dexcited to $n-1, n-2. n-3 , ......... 2,1.$ So total number of spectral lines in this case is$$ 1+2+3+......+(n-1)+ n = \frac {n(n+1)}{2}.$$ In your case $ n= n_2 - n_1$.