We are given the following two equations:

$P(n, r) = 240 \quad \text{and} \quad C(n, r) = 120$

We know the following formulas for permutations and combinations:

1. Permutation:
   $P(n, r) = \frac{n!}{(n - r)!}$
2. Combination:
   $C(n, r) = \frac{n!}{r!(n - r)!}$

To find the value of $r$, let's divide the permutation formula by the combination formula, which eliminates $n!$ and simplifies the equation:

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

Thus:

$\frac{240}{120} = r!$

Simplifying:

$2 = r!$

Now, we solve for $r$. The factorial of $r$ that equals 2 is:

$r = 2$

Thus, the value of $r$ is 2.

Would you like further details or explanations? Here are 5 related questions for further understanding:

1. How do you calculate $n!$ for any given $n$?
2. Can you explain why we divided $P(n, r)$ by $C(n, r)$?
3. What is the difference between permutations and combinations?
4. How does the factorial operation impact larger values of $r$?
5. How would the equations change if $P(n, r)$ and $C(n, r)$ were larger numbers?

Tip: The division trick for simplifying $P(n, r)$ and $C(n, r)$ is useful when you need to eliminate complex factorials.