Chris's pet tiger travels by jumping north and east. Chris wants to ride his tiger from Fine Hall to McCosh, which is $3$ jumps east and $10$ jumps north. However, Chris wants to avoid the horde of PUMaC competitors eating lunch at Frist, located $2$ jumps east and $4$ jumps north of Fine Hall. How many ways can he get to McCosh without going through Frist?
1. **Identify the coordinates and the problem setup:**
   - Fine Hall is at the origin \((0,0)\).
   - McCosh is at \((3,10)\).
   - Frist is at \((2,4)\).

2. **Calculate the total number of paths from Fine Hall to McCosh:**
   - To get from \((0,0)\) to \((3,10)\), Chris needs to make a total of \(3 + 10 = 13\) jumps, consisting of 3 east jumps and 10 north jumps.
   - The number of ways to arrange these jumps is given by the binomial coefficient:
     \[
     \binom{13}{3} = \frac{13!}{3! \cdot 10!}
     \]
   - Calculating this:
     \[
     \binom{13}{3} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286
     \]

3. **Calculate the number of paths that pass through Frist:**
   - First, calculate the number of ways to get from \((0,0)\) to \((2,4)\):
     \[
     \binom{6}{2} = \frac{6!}{2! \cdot 4!} = \frac{6 \times 5}{2 \times 1} = 15
     \]
   - Next, calculate the number of ways to get from \((2,4)\) to \((3,10)\):
     \[
     \binom{7}{1} = \frac{7!}{1! \cdot 6!} = 7
     \]
   - The total number of paths passing through \((2,4)\) is:
     \[
     15 \times 7 = 105
     \]

4. **Calculate the number of paths that do not pass through Frist:**
   - Subtract the number of paths passing through Frist from the total number of paths:
     \[
     286 - 105 = 181
     \]

Conclusion:
\[
\boxed{181}
\]