It seems like there might be a typo in your expression. I assume you meant to calculate $\frac{6!}{(6-3)! \cdot 3!}$. This expression represents a combination formula, specifically $\binom{6}{3}$.

Let's go through the calculation step-by-step.

Step 1: Rewrite the Expression

The expression is:

$\frac{6!}{(6-3)! \cdot 3!}$

Simplify the subtraction:

$6 - 3 = 3$

This gives:

$\frac{6!}{3! \cdot 3!}$

Step 2: Expand the Factorials

Now, let's expand the factorials:

$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$ $3! = 3 \times 2 \times 1 = 6$

Step 3: Cancel Out Common Terms

Rewrite $6!$ to make it easier to cancel out terms with $3!$:

$6! = (6 \times 5 \times 4) \times (3!)$

So, the expression becomes:

$\frac{(6 \times 5 \times 4) \times 3!}{3! \cdot 3!}$

The $3!$ in the numerator and one $3!$ in the denominator cancel each other out, leaving:

$\frac{6 \times 5 \times 4}{3!}$

Step 4: Compute the Remaining Expression

Compute the numerator:

$6 \times 5 = 30$ $30 \times 4 = 120$

Now divide by $3!$:

$3! = 6$

Thus, the expression simplifies to:

$\frac{120}{6} = 20$

Final Answer

The value of the expression $\frac{6!}{(6-3)! \cdot 3!}$ is:

$\boxed{20}$

Would you like more details or have any questions?

Here are 5 related questions to explore further:

1. How would you calculate $\frac{8!}{(8-4)! \cdot 4!}$?
2. What is the general formula for calculating $\binom{n}{r}$ (combinations)?
3. How are combinations different from permutations in terms of formula and usage?
4. How does the factorial function help in counting problems?
5. How can factorial expressions be used in probability theory?

Tip: To simplify factorial expressions, look for opportunities to cancel out common terms in the numerator and the denominator.