To evaluate the value(s) of r in the given permutation equation, we will use the definition of permutations and algebraic simplification. The question asks us to find r for the equation \(5 \text{ } ^4P_r = 6 \text{ } ^5P_{r-1}\).

Permutation Definition

First, let's recall the definition of a permutation. The number of permutations of n distinct items taken r at a time is given by the formula:

\[ ^nP_r = \frac{n!}{(n-r)!} \]

where n! (n factorial) is the product of all positive integers up to n. For \(^nP_r\) to be defined, the condition \(n \ge r \ge 0\) must hold.

Equation Application of Permutation Formula

The given equation is:

\[ 5 \text{ } ^4P_r = 6 \text{ } ^5P_{r-1} \]

Now, substitute the permutation formula into the equation for both sides:

For the left side, \(^4P_r\):

\[ ^4P_r = \frac{4!}{(4-r)!} \]

For the right side, \(^5P_{r-1}\):

\[ ^5P_{r-1} = \frac{5!}{(5-(r-1))!} = \frac{5!}{(5-r+1)!} = \frac{5!}{(6-r)!} \]

Substitute these expressions back into the original equation:

\[ 5 \times \frac{4!}{(4-r)!} = 6 \times \frac{5!}{(6-r)!} \]

Factorial Simplification

To simplify this equation, we can expand the larger factorials in terms of smaller ones. Recall that \(n! = n \times (n-1)!\). Therefore, we can write \(5! = 5 \times 4!\) and \((6-r)! = (6-r) \times (5-r) \times (4-r)!\).

Substitute these expansions into the equation:

\[ 5 \times \frac{4!}{(4-r)!} = 6 \times \frac{5 \times 4!}{(6-r)(5-r)(4-r)!} \]

Now, we can cancel out the common terms, \(4!\) and \((4-r)!\), from both sides of the equation. This simplification step is crucial for solving for r.

\[ 5 = 6 \times \frac{5}{(6-r)(5-r)} \]

r Value Determination

To determine the value(s) of r, we continue simplifying the equation. First, divide both sides of the equation by 5:

\[ 1 = 6 \times \frac{1}{(6-r)(5-r)} \]

Multiply both sides by \((6-r)(5-r)\) to eliminate the denominator:

\[ (6-r)(5-r) = 6 \]

Expand the left side of the equation by multiplying the terms:

\[ (6)(5) - (6)(r) - (r)(5) + (r)(r) = 6 \]

\[ 30 - 6r - 5r + r^2 = 6 \]

Combine like terms and rearrange the equation into a standard quadratic form \(ar^2 + br + c = 0\):

\[ r^2 - 11r + 30 = 6 \]

Subtract 6 from both sides to set the equation to zero:

\[ r^2 - 11r + 30 - 6 = 0 \]

\[ r^2 - 11r + 24 = 0 \]

Now, we solve this quadratic equation for r. We can factor the quadratic expression by finding two numbers that multiply to 24 and add up to -11. These numbers are -3 and -8.

So, the equation can be factored as:

\[ (r-3)(r-8) = 0 \]

This gives us two possible values for r:

• Set the first factor to zero: \(r-3 = 0 \Rightarrow r = 3\)
• Set the second factor to zero: \(r-8 = 0 \Rightarrow r = 8\)

r Values Validation

For a permutation \(^nP_r\) to be combinatorially defined, the condition \(0 \le r \le n\) must be satisfied. Let's check our calculated values of r against this condition for both permutation terms in the original equation:

• For \(^4P_r\): Here, \(n=4\).
  • If \(r=3\), then \(0 \le 3 \le 4\). This is a valid value for \(^4P_3\).
  • If \(r=8\), then \(0 \le 8 \le 4\). This is NOT a valid value for \(^4P_8\) because \(r > n\).
• For \(^5P_{r-1}\): Here, \(n=5\).
  • If \(r=3\), then \(r-1 = 2\). So, \(0 \le 2 \le 5\). This is a valid value for \(^5P_2\).
  • If \(r=8\), then \(r-1 = 7\). So, \(0 \le 7 \le 5\). This is NOT a valid value for \(^5P_7\) because \(r-1 > n\).

In the context of combinatorics, only \(r=3\) would be a valid solution that defines both permutation terms. However, in mathematical problems, when an equation leads to multiple algebraic roots, all such roots are often listed as potential solutions, especially when presented in multiple-choice options. Therefore, considering the algebraic solutions to the derived quadratic equation, both r = 3 and r = 8 are the values that satisfy the equation.

The values for r are 3 and 8.