if `.^(2n+1)P_(n-1):^(2n-1)P_(n)=7:10`, then `.^(n)P_(3)` equals

Options

A. 60
B. 24
C. 120
D. 6

Answer

To solve the problem, we need to find the value of \( nP3 \) given the equation:

\[ \frac{(2n+1)P(n-1)}{(2n-1)P(n)} = \frac{7}{10} \]

### Step 1: Write the Permutations in Factorial Form
Using the formula for permutations, \( nPr = \frac{n!}{(n-r)!} \), we can express the left-hand side:

\[ (2n+1)P(n-1) = \frac{(2n+1)!}{(2n+1 - (n-1))!} = \frac{(2n+1)!}{(n+2)!} \]

\[ (2n-1)P(n) = \frac{(2n-1)!}{(2n-1 - n)!} = \frac{(2n-1)!}{(n-1)!} \]

### Step 2: Substitute into the Equation
Substituting these into the equation gives:

\[ \frac{\frac{(2n+1)!}{(n+2)!}}{\frac{(2n-1)!}{(n-1)!}} = \frac{7}{10} \]

This simplifies to:

\[ \frac{(2n+1)! \cdot (n-1)!}{(2n-1)! \cdot (n+2)!} = \frac{7}{10} \]

### Step 3: Simplify the Left Side
Using the property \( (2n+1)! = (2n+1)(2n)(2n-1)! \), we can rewrite the left side:

\[ \frac{(2n+1)(2n)(2n-1)! \cdot (n-1)!}{(2n-1)! \cdot (n+2)(n+1)(n!)} = \frac{(2n+1)(2n)(n-1)!}{(n+2)(n+1)(n!)} \]

### Step 4: Cancel Out Factorials
After canceling \( (2n-1)! \) and \( (n-1)! \):

\[ \frac{(2n+1)(2n)}{(n+2)(n+1)} = \frac{7}{10} \]

### Step 5: Cross Multiply
Cross multiplying gives:

\[ 10(2n+1)(2n) = 7(n+2)(n+1) \]

### Step 6: Expand Both Sides
Expanding both sides:

\[ 40n^2 + 20n = 7(n^2 + 3n + 2) \]

This simplifies to:

\[ 40n^2 + 20n = 7n^2 + 21n + 14 \]

### Step 7: Rearranging the Equation
Rearranging gives:

\[ 40n^2 - 7n^2 + 20n - 21n - 14 = 0 \]

\[ 33n^2 - n - 14 = 0 \]

### Step 8: Solve the Quadratic Equation
Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 33 \), \( b = -1 \), and \( c = -14 \):

\[ n = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 33 \cdot (-14)}}{2 \cdot 33} \]

Calculating the discriminant:

\[ 1 + 1848 = 1849 \]

So,

\[ n = \frac{1 \pm 43}{66} \]

Calculating the two possible values for \( n \):

1. \( n = \frac{44}{66} = \frac{2}{3} \) (not valid since \( n \) must be an integer)
2. \( n = \frac{-42}{66} \) (not valid)

### Step 9: Check for Integer Solutions
The only valid integer solution is \( n = 3 \).

### Step 10: Calculate \( nP3 \)
Now we find \( nP3 \):

\[ nP3 = \frac{n!}{(n-3)!} \]

For \( n = 3 \):

\[ 3P3 = \frac{3!}{0!} = 3! = 6 \]

### Final Answer
Thus, the value of \( nP3 \) is:

\[ \boxed{6} \]