it is given that the permutation P(n, r) = 240 and combination C(n,r) = 120 find the value of r



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.

it is given that the permutation P(n, r) = 240 and combination C(n,r) = 120 find the value of r

Learn how to solve for r when given the permutation P(n, r) = 240 and combination C(n, r) = 120. This problem involves understanding permutations, combinations, and factorials. Suitable for students in Grades 9-11.

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