The image shows the following equation:

$4P(n + 2, r + 2) = 20P(n + 1, r + 1)$

where $P(n, r)$ typically represents the permutation formula, defined as:

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

Let's simplify and solve this equation.

Step-by-Step Solution:

1. Write Down the Permutation Expressions:

   • $P(n + 2, r + 2) = \frac{(n + 2)!}{(n + 2 - (r + 2))!} = \frac{(n + 2)!}{(n - r)!}$
   • $P(n + 1, r + 1) = \frac{(n + 1)!}{(n + 1 - (r + 1))!} = \frac{(n + 1)!}{(n - r)!}$

2. Substitute and Set the Equation: $4 \cdot \frac{(n + 2)!}{(n - r)!} = 20 \cdot \frac{(n + 1)!}{(n - r)!}$

3. Cancel Out the Common Denominator: Since both sides have $(n - r)!$, cancel it out: $4(n + 2)! = 20(n + 1)!$

4. Expand the Factorials:

   • $(n + 2)! = (n + 2)(n + 1)!$

   Substitute to get: $4(n + 2)(n + 1)! = 20(n + 1)!$

5. Divide Both Sides by $(n + 1)!$: $4(n + 2) = 20$

6. Solve for $n$: $n + 2 = \frac{20}{4} = 5$ $n = 3$

Thus, the solution is: $n = 3$

Would you like a more detailed explanation of any of these steps, or do you have any other questions?

Related Questions:

1. What is the permutation formula and how is it derived?
2. Can we solve a similar problem using combinations instead of permutations?
3. What happens if we change the coefficients on both sides of the equation?
4. How would the solution change if we used $P(n, r) = n^r$ instead?
5. Can this equation be solved if $n$ is a non-integer?

Tip:

When working with permutations, always ensure to expand the factorials step-by-step to avoid common simplification errors.