Let $n$ be the number of ordered quadruples $(x_1,x_2,x_3,x_4)$ of positive odd [integers](https://artofproblemsolving.com/wiki/index.php/Integer) that satisfy $\sum_{i = 1}^4 x_i = 98.$  Find $\frac n{100}.$
We want $x_1 +x_2+x_3+x_4 =98$. This seems like it can be solved with stars and bars, however note that the quadruples all need to be odd. This motivates us to set $x_i= 2y_i +1$, as for all integers $y_i$, $2y_i + 1$ will be odd. Substituting we get \[2y_1+2y_2+2y_3+2y_4 +4 = 98 \implies y_1+y_2+y_3+y_4 =47\]
Note that this is an algebraic bijection, we have simplified the problem and essentially removed the odd condition, so now we can finish with plain stars as bars, which gives us $n= {50\choose3}$. Computing this and dividing by 100 gives us an answer of $\boxed{196}$.
~smartguy888