Define it!

Since $x_1,x_2,x_3,x_4$ are odd, let $x_1=\color{#0C6AC7}{2p+1},x_2=\color{#D61F06}{2q+1},x_3=\color{#456461}{2r+1},x_4=\color{#EC7300}{2s+1}$ for some integers p,q,r,s.
Substituting them in the given expression, given equation simplifies to: $\color{#0C6AC7}{p}+\color{#D61F06}{q}+\color{#456461}{r}+\color{#EC7300}{s}=47$ From here it's a regular combinatorics problem asking for non negative integers $p,q,r,s$ satisfying the above equation, which are given by $\binom{47+4-1}{47}=\binom{50}{47}$ $\Large\therefore \dfrac{\binom{50}{47}}{100}=\huge\color{#302B94}{\boxed{\color{#007fff}{196}}}$