To solve the problem, we need to understand the notation and the concept of permutations. The notation \( ^nP_r \) represents the number of ways to arrange \( r \) objects from a total of \( n \) distinct objects. The formula for permutations is given by: \[ ^nP_r = \frac{n!}{(n-r)!} \] where \( n! \) (n factorial) is the product of all positive integers up to \( n \), and \( (n-r)! \) is the factorial of the difference between \( n \) and \( r \). ### Step 1: Understanding the Given Information We are given that: \[ ^6P_r = 6 \] Using the formula for permutations, we can express this as: \[ ^6P_r = \frac{6!}{(6-r)!} = 6 \] ### Step 2: Simplifying the Equation Now, we can simplify \( 6! \): \[ 6! = 720 \] So, we can rewrite the equation: \[ \frac{720}{(6-r)!} = 6 \] ### Step 3: Solving for \( (6-r)! \) To isolate \( (6-r)! \), we multiply both sides by \( (6-r)! \): \[ 720 = 6 \cdot (6-r)! \] Now, divide both sides by 6: \[ (6-r)! = \frac{720}{6} = 120 \] ### Step 4: Finding \( 6-r \) Next, we need to find \( 6-r \) such that \( (6-r)! = 120 \). We know that: \[ 5! = 120 \] Thus, we have: \[ 6 - r = 5 \] From this, we can solve for \( r \): \[ r = 6 - 5 = 1 \] ### Step 5: Finding \( ^6P_{r+1} \) Now that we have \( r = 1 \), we need to find \( ^6P_{r+1} \), which is \( ^6P_2 \): \[ ^6P_2 = \frac{6!}{(6-2)!} = \frac{6!}{4!} \] Calculating \( 6! \) and \( 4! \): \[ 6! = 720 \quad \text{and} \quad 4! = 24 \] Now substituting these values into the permutation formula: \[ ^6P_2 = \frac{720}{24} = 30 \] ### Conclusion Thus, the value of \( ^6P_{r+1} \) is: \[ \boxed{30} \] ### Explanation of Other Options - **Option B (33)**: This is incorrect because it does not match the calculated value of \( ^6P_2 \). - **Option C (35)**: This is also incorrect as it does not correspond to any permutation calculation for \( n = 6 \) and \( r = 2 \). - **Option D (15)**: This is incorrect as well; it does not fit the permutation formula for the given values. ### Revision Summary - The formula for permutations is \( ^nP_r = \frac{n!}{(n-r)!} \). - We found \( r \) by solving \( ^6P_r = 6 \) and determined \( r = 1 \). - We calculated \( ^6P_{r+1} = ^6P_2 = 30 \). - The correct answer is \( \boxed{30} \).