If `""^(11)P_(r )= ""^(12)P_(r-1)`, then r = ………….

Answer

To solve the equation \( ^{11}P_r = ^{12}P_{r-1} \), we will use the formula for permutations. The formula for permutations is given by:

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

### Step 1: Write the permutations using the formula

Using the formula, we can express both sides of the equation:

\[ ^{11}P_r = \frac{11!}{(11-r)!} \]
\[ ^{12}P_{r-1} = \frac{12!}{(12 - (r-1))!} = \frac{12!}{(13 - r)!} \]

### Step 2: Set the two expressions equal to each other

Now we can set the two expressions equal to each other:

\[ \frac{11!}{(11-r)!} = \frac{12!}{(13-r)!} \]

### Step 3: Simplify the right side

We know that \( 12! = 12 \times 11! \), so we can substitute that into the equation:

\[ \frac{11!}{(11-r)!} = \frac{12 \times 11!}{(13-r)!} \]

### Step 4: Cancel \( 11! \) from both sides

Since \( 11! \) appears on both sides, we can cancel it out:

\[ \frac{1}{(11-r)!} = \frac{12}{(13-r)!} \]

### Step 5: Cross-multiply to eliminate the fractions

Cross-multiplying gives us:

\[ (13-r)! = 12 \times (11-r)! \]

### Step 6: Expand \( (13-r)! \)

We can expand \( (13-r)! \) as follows:

\[ (13-r)(12-r)(11-r)! = 12 \times (11-r)! \]

### Step 7: Cancel \( (11-r)! \)

Since \( (11-r)! \) appears on both sides, we can cancel it out:

\[ (13-r)(12-r) = 12 \]

### Step 8: Expand and rearrange the equation

Expanding the left side gives us:

\[ 156 - 25r + r^2 = 12 \]

Rearranging gives:

\[ r^2 - 25r + 144 = 0 \]

### Step 9: Factor the quadratic equation

Now we can factor the quadratic equation:

\[ (r - 16)(r - 9) = 0 \]

### Step 10: Solve for \( r \)

Setting each factor to zero gives us:

\[ r - 16 = 0 \quad \text{or} \quad r - 9 = 0 \]

Thus, we find:

\[ r = 16 \quad \text{or} \quad r = 9 \]

### Step 11: Determine the valid solution

Since \( r \) must be less than or equal to 12 (as \( n = 12 \)), we discard \( r = 16 \) and accept:

\[ r = 9 \]

### Final Answer

Thus, the value of \( r \) is:

\[ \boxed{9} \]