To evaluate the expression:

$\frac{C(6, 4) \cdot C(7, 5)}{C(13, 9)}$

We first calculate each combination separately using the formula for combinations:

$C(n, r) = \frac{n!}{r!(n-r)!}$

1. Calculate $C(6, 4)$:

   $C(6, 4) = \frac{6!}{4!(6-4)!} = \frac{6!}{4! \cdot 2!} = \frac{720}{24 \cdot 2} = \frac{720}{48} = 15$

2. Calculate $C(7, 5)$:

   $C(7, 5) = \frac{7!}{5!(7-5)!} = \frac{7!}{5! \cdot 2!} = \frac{5040}{120 \cdot 2} = \frac{5040}{240} = 21$

3. Calculate $C(13, 9)$:

   $C(13, 9) = \frac{13!}{9!(13-9)!} = \frac{13!}{9! \cdot 4!} = \frac{6227020800}{362880 \cdot 24} = \frac{6227020800}{8709120} = 715$

Now, substitute the calculated values into the original expression:

$\frac{C(6, 4) \cdot C(7, 5)}{C(13, 9)} = \frac{15 \cdot 21}{715} = \frac{315}{715} = \frac{63}{143}$

The expression simplifies to $\frac{63}{143}$.

Would you like more details on any part of the calculation or have any other questions?

---

Here are 5 related questions:

1. How do you calculate the number of permutations for a given set?
2. What is the difference between combinations and permutations?
3. How would the expression change if we used $C(6, 3)$ instead of $C(6, 4)$?
4. Can you express the original combination problem in a different context?
5. What is the binomial theorem, and how do combinations relate to it?

Tip: When solving combination problems, double-check the factorial calculations, especially for large numbers, to avoid mistakes.