As, $n_1 \geq 1, n_2 \geq 2, n_3 \geq 3, n_4 \geq 4, n_5 \geq 5$

Let $n_1-1=x_1 \geq 0, n_2-2=x_2 \geq 0, \ldots$, $n_5-5=x_5 \geq 0$

$\Rightarrow$ New equation will be

$$ \begin{gathered} x_1+1+x_2+2+\ldots+x_5+5=20 \end{gathered} $$

$\begin{array}{r}\Rightarrow x_1+x_2+x_3+x_4+x_5 =20-15=5\end{array}$

Now, $ x_1 \leq x_2 \leq x_3 \leq x_4 \leq x_5$

So the cases can be listed giving -

$$ \begin{array}{c|c|c|c|c} \hline \boldsymbol{x}_{\boldsymbol{1}} & \boldsymbol{x}_{\mathbf{2}} & \boldsymbol{x}_{\mathbf{3}} & \boldsymbol{x}_{\mathbf{4}} & \boldsymbol{x}_{\mathbf{5}} \\ \hline 0 & 0 & 0 & 0 & 5 \\\hline 0 & 0 & 0 & 1 & 4 \\\hline 0 & 0 & 0 & 2 & 3 \\\hline 0 & 0 & 1 & 1 & 3 \\\hline 0 & 0 & 1 & 2 & 2 \\\hline 0 & 1 & 1 & 1 & 2 \\\hline 1 & 1 & 1 & 1 & 1 \\ \hline \end{array} $$

So there are 7 distinct arrangements.