Find n, ` ""^(n+5)P_(n+1)=(11)/(2)(n-1)*""^(n+3)P_(n).`

Answer

To solve the equation \( (n+5)P(n+1) = \frac{11}{2}(n-1)(n+3)P(n) \), we will follow these steps:

### Step 1: Write the permutations in factorial form
The permutation \( nPr \) can be expressed as:
\[ nPr = \frac{n!}{(n-r)!} \]
Thus, we can rewrite the left-hand side:
\[ (n+5)P(n+1) = \frac{(n+5)!}{(n+5-(n+1))!} = \frac{(n+5)!}{(4)!} \]
And for the right-hand side:
\[ (n+3)P(n) = \frac{(n+3)!}{(n+3-n)!} = \frac{(n+3)!}{(3)!} \]
So, we can rewrite the equation as:
\[ \frac{(n+5)!}{4!} = \frac{11}{2}(n-1) \cdot \frac{(n+3)!}{3!} \]

### Step 2: Simplify the equation
Substituting \( 4! = 24 \) and \( 3! = 6 \):
\[ \frac{(n+5)!}{24} = \frac{11}{2}(n-1) \cdot \frac{(n+3)!}{6} \]
Multiplying both sides by 24 to eliminate the fraction:
\[ (n+5)! = 11(n-1)(n+3)! \]

### Step 3: Express \( (n+5)! \) in terms of \( (n+3)! \)
Using the property of factorials:
\[ (n+5)! = (n+5)(n+4)(n+3)! \]
Substituting this back into the equation:
\[ (n+5)(n+4)(n+3)! = 11(n-1)(n+3)! \]
We can cancel \( (n+3)! \) from both sides (assuming \( n \neq -3 \)):
\[ (n+5)(n+4) = 11(n-1) \]

### Step 4: Expand and rearrange the equation
Expanding both sides:
\[ n^2 + 9n + 20 = 11n - 11 \]
Rearranging gives:
\[ n^2 + 9n + 20 - 11n + 11 = 0 \]
This simplifies to:
\[ n^2 - 2n + 31 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -2, c = 31 \):
\[ n = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 31}}{2 \cdot 1} \]
Calculating the discriminant:
\[ n = \frac{2 \pm \sqrt{4 - 124}}{2} \]
\[ n = \frac{2 \pm \sqrt{-120}}{2} \]
Since the discriminant is negative, there are no real solutions for \( n \).

### Conclusion
The equation has no real solutions for \( n \).