Find r if 5 P(4,r) = 6P ( 5,r-1) , ` r ge 1.`

Answer

To solve the equation \( 5 P(4,r) = 6 P(5,r-1) \), we will follow these steps:

### Step 1: Write the permutation formulas
The formula for permutations is given by:
\[ nPr = \frac{n!}{(n-r)!} \]
So, we can express \( 4Pr \) and \( 5P(r-1) \) using this formula.

### Step 2: Substitute the permutation formulas into the equation
Substituting the formulas into the original equation:
\[ 5 \cdot \frac{4!}{(4-r)!} = 6 \cdot \frac{5!}{(5-(r-1))!} \]
This simplifies to:
\[ 5 \cdot \frac{4!}{(4-r)!} = 6 \cdot \frac{5!}{(6-r)!} \]

### Step 3: Simplify the right-hand side
We know that \( 5! = 5 \cdot 4! \), so we can rewrite the right-hand side:
\[ 5 \cdot \frac{4!}{(4-r)!} = 6 \cdot \frac{5 \cdot 4!}{(6-r)!} \]
Now, we can cancel \( 4! \) from both sides:
\[ 5 \cdot \frac{1}{(4-r)!} = 30 \cdot \frac{1}{(6-r)!} \]

### Step 4: Cross-multiply to eliminate the fractions
Cross-multiplying gives us:
\[ 5 \cdot (6-r)! = 30 \cdot (4-r)! \]

### Step 5: Expand the factorials
We can expand \( (6-r)! \) and \( (4-r)! \):
\[ (6-r)! = (6-r)(5-r)(4-r)! \]
Substituting this back into the equation:
\[ 5 \cdot (6-r)(5-r)(4-r)! = 30 \cdot (4-r)! \]

### Step 6: Cancel \( (4-r)! \)
Assuming \( (4-r)! \neq 0 \) (which is valid since \( r \leq 4 \)), we can cancel \( (4-r)! \):
\[ 5 \cdot (6-r)(5-r) = 30 \]

### Step 7: Simplify the equation
Dividing both sides by 5:
\[ (6-r)(5-r) = 6 \]

### Step 8: Expand and rearrange
Expanding the left side:
\[ 30 - 11r + r^2 = 6 \]
Rearranging gives us:
\[ r^2 - 11r + 24 = 0 \]

### Step 9: Factor the quadratic equation
Factoring the quadratic:
\[ (r - 8)(r - 3) = 0 \]
Thus, the solutions for \( r \) are:
\[ r = 8 \quad \text{or} \quad r = 3 \]

### Step 10: Apply the condition \( r \geq 1 \)
Since \( r \) must be greater than or equal to 1, both solutions are valid. However, since \( r \) must also be less than or equal to 4 (as it is a permutation of 4), we have:
\[ r = 3 \]

### Final Answer
Thus, the value of \( r \) is:
\[ \boxed{3} \]