SOLUTION —
$\begin{array}{l}\text { }\frac{1}{1 !(n-1) !}+\frac{1}{3 !(n-3) !}+\frac{1}{5 !(n-5) !}+\ldots \\=\frac{1}{n !}\left[\frac{1}{1 !} \cdot \frac{n !}{(n-1) !}+\frac{1}{3 !} \cdot \frac{n !}{(n-3) !}+\frac{1}{5 !} \cdot \frac{n !}{(n-5) !}+\ldots\right] \\=\frac{1}{n !}\left[{ }^n C_1+{ }^n C_3+{ }^n C_5+\ldots\right] \\=\frac{2^{n-1}}{n !} \\\end{array}$
So, The correct option will be (B).