If `""^12 P_r=""^11P_6+6."^11P_5` then is r equal to

Options

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

Answer

To solve the equation \( ^{12}P_r = ^{11}P_6 + 6 \cdot ^{11}P_5 \), we will use the formula for permutations, which is given by:

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

### Step 1: Write the equation using the permutation formula
We start by substituting the permutation formulas into the equation:

\[ ^{12}P_r = \frac{12!}{(12-r)!} \]
\[ ^{11}P_6 = \frac{11!}{(11-6)!} = \frac{11!}{5!} \]
\[ ^{11}P_5 = \frac{11!}{(11-5)!} = \frac{11!}{6!} \]

Now substituting these into the original equation, we get:

\[ \frac{12!}{(12-r)!} = \frac{11!}{5!} + 6 \cdot \frac{11!}{6!} \]

### Step 2: Simplify the right-hand side
We can factor out \( 11! \) from the right-hand side:

\[ \frac{12!}{(12-r)!} = 11! \left( \frac{1}{5!} + \frac{6}{6!} \right) \]

Now, simplifying \( \frac{6}{6!} \):

\[ \frac{6}{6!} = \frac{6}{6 \cdot 5!} = \frac{1}{5!} \]

Thus, the right-hand side becomes:

\[ \frac{12!}{(12-r)!} = 11! \left( \frac{1}{5!} + \frac{1}{5!} \right) = 11! \cdot \frac{2}{5!} \]

### Step 3: Rewrite \( 12! \) in terms of \( 11! \)
We know that:

\[ 12! = 12 \cdot 11! \]

Substituting this into our equation gives:

\[ \frac{12 \cdot 11!}{(12-r)!} = 11! \cdot \frac{2}{5!} \]

### Step 4: Cancel \( 11! \) from both sides
Assuming \( 11! \neq 0 \), we can cancel \( 11! \):

\[ \frac{12}{(12-r)!} = \frac{2}{5!} \]

### Step 5: Cross-multiply to solve for \( (12-r)! \)
Cross-multiplying gives:

\[ 12 \cdot 5! = 2 \cdot (12-r)! \]

Calculating \( 5! \):

\[ 5! = 120 \]

So we have:

\[ 12 \cdot 120 = 2 \cdot (12-r)! \]

This simplifies to:

\[ 1440 = 2 \cdot (12-r)! \]

Dividing both sides by 2:

\[ 720 = (12-r)! \]

### Step 6: Find \( r \)
We know that \( 720 = 6! \), thus:

\[ (12-r)! = 6! \]

This implies:

\[ 12 - r = 6 \implies r = 12 - 6 = 6 \]

### Final Answer
Thus, the value of \( r \) is:

\[ \boxed{6} \]