What is the value of r, if P ( 5,r) = P ( 6,r-1) ?

Options

A. 9
B. 5
C. 4
D. 2

Answer

To solve the equation \( P(5, r) = P(6, r-1) \), we will use the formula for permutations, which is given by:

\[ P(n, r) = \frac{n!}{(n-r)!} \]

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

\[ P(5, r) = \frac{5!}{(5-r)!} \]
\[ P(6, r-1) = \frac{6!}{(6-(r-1))!} = \frac{6!}{(7-r)!} \]

So, we can rewrite the equation as:

\[ \frac{5!}{(5-r)!} = \frac{6!}{(7-r)!} \]

### Step 2: Substitute the factorial values
Now, we know that \( 6! = 6 \times 5! \). Thus, we can substitute this into our equation:

\[ \frac{5!}{(5-r)!} = \frac{6 \times 5!}{(7-r)!} \]

### Step 3: Cancel \( 5! \) from both sides
Since \( 5! \) is common on both sides, we can cancel it out:

\[ \frac{1}{(5-r)!} = \frac{6}{(7-r)!} \]

### Step 4: Cross-multiply to eliminate the fractions
Cross-multiplying gives us:

\[ (7-r)! = 6 \times (5-r)! \]

### Step 5: Expand the factorials
We can expand \( (7-r)! \):

\[ (7-r)(6-r)(5-r)! = 6 \times (5-r)! \]

### Step 6: Cancel \( (5-r)! \) from both sides
Again, we can cancel \( (5-r)! \) from both sides (assuming \( r \neq 5 \)):

\[ (7-r)(6-r) = 6 \]

### Step 7: Expand and rearrange the equation
Expanding the left side gives:

\[ 42 - 13r + r^2 = 6 \]

Rearranging this, we have:

\[ r^2 - 13r + 36 = 0 \]

### Step 8: Factor the quadratic equation
Now, we will factor the quadratic equation:

\[ (r - 9)(r - 4) = 0 \]

### Step 9: Solve for \( r \)
Setting each factor to zero gives us the possible values for \( r \):

\[ r - 9 = 0 \quad \Rightarrow \quad r = 9 \]
\[ r - 4 = 0 \quad \Rightarrow \quad r = 4 \]

### Step 10: Determine the valid value of \( r \)
Since \( r \) must be less than or equal to \( n \) (where \( n \) is 5 or 6), \( r = 9 \) is not valid. Thus, the only valid solution is:

\[ r = 4 \]

### Final Answer
The value of \( r \) is \( 4 \).

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