If `.^(n)P_(3)+.^(n)C_(n-2)=14n`, the value of n is
(a) 5 (b) 6 (c) 8 (d) 10

Options

A. 5
B. 6
C. 8
D. 10

Answer

To solve the equation \( ^nP_3 + ^nC_{n-2} = 14n \), we will break it down into steps.

### Step 1: Write the formulas for permutations and combinations
The permutation \( ^nP_3 \) is given by:
\[ ^nP_3 = \frac{n!}{(n-3)!} = n(n-1)(n-2) \]
The combination \( ^nC_{n-2} \) can be rewritten using the property \( ^nC_{r} = ^nC_{n-r} \):
\[ ^nC_{n-2} = ^nC_2 = \frac{n!}{2!(n-2)!} = \frac{n(n-1)}{2} \]

### Step 2: Substitute the formulas into the equation
Now substituting these into the original equation:
\[ n(n-1)(n-2) + \frac{n(n-1)}{2} = 14n \]

### Step 3: Eliminate \( n \) from both sides
We can factor out \( n(n-1) \) from the left-hand side:
\[ n(n-1) \left( (n-2) + \frac{1}{2} \right) = 14n \]
Dividing both sides by \( n \) (assuming \( n \neq 0 \)):
\[ (n-1) \left( (n-2) + \frac{1}{2} \right) = 14 \]

### Step 4: Simplify the equation
Now simplify the expression inside the parentheses:
\[ (n-2) + \frac{1}{2} = n - \frac{4}{2} + \frac{1}{2} = n - \frac{3}{2} \]
Thus, we have:
\[ (n-1) \left( n - \frac{3}{2} \right) = 14 \]

### Step 5: Expand and rearrange the equation
Expanding the left-hand side:
\[ n^2 - \frac{3}{2}n - n + \frac{3}{2} = 14 \]
This simplifies to:
\[ n^2 - \frac{5}{2}n + \frac{3}{2} - 14 = 0 \]
Multiplying through by 2 to eliminate the fraction:
\[ 2n^2 - 5n + 3 - 28 = 0 \]
Thus:
\[ 2n^2 - 5n - 25 = 0 \]

### Step 6: Factor the quadratic equation
Now we will factor the quadratic:
\[ 2n^2 - 10n + 5n - 25 = 0 \]
Factoring by grouping:
\[ 2n(n - 5) + 5(n - 5) = 0 \]
This gives:
\[ (n - 5)(2n + 5) = 0 \]

### Step 7: Solve for \( n \)
Setting each factor to zero gives:
1. \( n - 5 = 0 \) → \( n = 5 \)
2. \( 2n + 5 = 0 \) → \( n = -\frac{5}{2} \) (not valid as \( n \) must be a positive integer)

Thus, the only valid solution is:
\[ n = 5 \]

### Conclusion
The value of \( n \) is \( 5 \).