If `""^(2n+1)P_(n-1) : ""^(2n-1)P_(n) =3 :5 `the possible value of n will be :

Options

A. 3
B. 5
C. 4
D. 2

Answer

To solve the equation \(\frac{(2n+1)P(n-1)}{(2n-1)P(n)} = \frac{3}{5}\), we start by expressing the permutations in terms of factorials.

### Step 1: Write the permutations in factorial form
The permutation \(nPr\) is given by the formula:
\[ nPr = \frac{n!}{(n-r)!} \]
So, we can write:
\[ (2n+1)P(n-1) = \frac{(2n+1)!}{(2n+1 - (n-1))!} = \frac{(2n+1)!}{(n+2)!} \]
and
\[ (2n-1)P(n) = \frac{(2n-1)!}{(2n-1 - n)!} = \frac{(2n-1)!}{(n-1)!} \]

### Step 2: Substitute these into the ratio
Now substituting these into the ratio:
\[ \frac{\frac{(2n+1)!}{(n+2)!}}{\frac{(2n-1)!}{(n-1)!}} = \frac{(2n+1)! \cdot (n-1)!}{(2n-1)! \cdot (n+2)!} \]

### Step 3: Simplify the expression
Now we can simplify the expression:
\[ \frac{(2n+1)(2n)(2n-1)! \cdot (n-1)!}{(2n-1)! \cdot (n+2)(n+1)(n!)} = \frac{(2n+1)(2n)}{(n+2)(n+1)} \]

### Step 4: Set the ratio equal to 3/5
Now we set the simplified expression equal to \(\frac{3}{5}\):
\[ \frac{(2n+1)(2n)}{(n+2)(n+1)} = \frac{3}{5} \]

### Step 5: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:
\[ 5(2n+1)(2n) = 3(n+2)(n+1) \]

### Step 6: Expand both sides
Expanding both sides:
\[ 5(4n^2 + 2n) = 3(n^2 + 3n + 2) \]
\[ 20n^2 + 10n = 3n^2 + 9n + 6 \]

### Step 7: Rearrange the equation
Rearranging gives:
\[ 20n^2 - 3n^2 + 10n - 9n - 6 = 0 \]
\[ 17n^2 + n - 6 = 0 \]

### Step 8: Factor the quadratic equation
Now we can factor or use the quadratic formula to solve for \(n\):
Using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 17\), \(b = 1\), and \(c = -6\):
\[ n = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 17 \cdot (-6)}}{2 \cdot 17} \]
\[ n = \frac{-1 \pm \sqrt{1 + 408}}{34} \]
\[ n = \frac{-1 \pm \sqrt{409}}{34} \]

### Step 9: Calculate the possible values of \(n\)
Calculating the square root and simplifying gives us the possible values of \(n\). Since \(n\) must be a positive integer, we evaluate the results.

### Final Result
The possible values of \(n\) can be calculated from the above expression.