If `""^(12)P_(r)=""^(11)P_(6)+6""^(11)P_(5)` then r is equal to

Options

A. 6
B. 5
C. 7
D. none of these

Answer

To solve the equation \( 12P_r = 11P_6 + 6 \cdot 11P_5 \), we will follow these steps:

### Step 1: Write down the given equation
We start with the equation:
\[ 12P_r = 11P_6 + 6 \cdot 11P_5 \]

### Step 2: Use the formula for permutations
Recall the formula for permutations:
\[ nP_r = \frac{n!}{(n-r)!} \]
Using this, we can express \( 11P_6 \) and \( 11P_5 \):
\[ 11P_6 = \frac{11!}{(11-6)!} = \frac{11!}{5!} \]
\[ 11P_5 = \frac{11!}{(11-5)!} = \frac{11!}{6!} \]

### Step 3: Substitute the values into the equation
Substituting these values back into the equation gives:
\[ 12P_r = \frac{11!}{5!} + 6 \cdot \frac{11!}{6!} \]

### Step 4: Simplify the right-hand side
Now, we simplify the right-hand side:
\[ 6 \cdot \frac{11!}{6!} = \frac{6 \cdot 11!}{6!} = \frac{11!}{5!} \]
Thus, we can write:
\[ 12P_r = \frac{11!}{5!} + \frac{11!}{5!} = 2 \cdot \frac{11!}{5!} \]

### Step 5: Express \( 12P_r \) using the permutation formula
Now, we can express \( 12P_r \) using the permutation formula:
\[ 12P_r = \frac{12!}{(12-r)!} \]

### Step 6: Set the two expressions equal to each other
We now have:
\[ \frac{12!}{(12-r)!} = 2 \cdot \frac{11!}{5!} \]

### Step 7: Simplify the left-hand side
We can express \( 12! \) as \( 12 \cdot 11! \):
\[ \frac{12 \cdot 11!}{(12-r)!} = 2 \cdot \frac{11!}{5!} \]
Cancelling \( 11! \) from both sides gives:
\[ \frac{12}{(12-r)!} = \frac{2}{5!} \]

### Step 8: Cross-multiply to solve for \( r \)
Cross-multiplying yields:
\[ 12 \cdot 5! = 2 \cdot (12-r)! \]
Calculating \( 5! = 120 \):
\[ 12 \cdot 120 = 2 \cdot (12-r)! \]
\[ 1440 = 2 \cdot (12-r)! \]
Dividing both sides by 2 gives:
\[ 720 = (12-r)! \]

### Step 9: Find \( r \) by determining \( (12-r) \)
We know \( 720 = 6! \), so:
\[ 12 - r = 6 \]
Thus, solving for \( r \):
\[ r = 12 - 6 = 6 \]

### Final Answer
Therefore, the value of \( r \) is:
\[ \boxed{6} \]