If `.^(n)P_(5)=20 .^(n)P_(3)`, find the value of n.

Explanation

To solve the equation \(.^{(n)}P_{5} = 20 .^{(n)}P_{3}\), we will follow these steps:

### Step 1: Write the Permutation Formula
The formula for permutations is given by:
\[ ^{(n)}P_r = \frac{n!}{(n-r)!} \]
Using this, we can express \(.^{(n)}P_{5}\) and \(.^{(n)}P_{3}\):
\[ ^{(n)}P_{5} = \frac{n!}{(n-5)!} \]
\[ ^{(n)}P_{3} = \frac{n!}{(n-3)!} \]

### Step 2: Set Up the Equation
Substituting these expressions into the original equation:
\[ \frac{n!}{(n-5)!} = 20 \cdot \frac{n!}{(n-3)!} \]

### Step 3: Cancel \(n!\)
Since \(n!\) appears in both sides of the equation, we can cancel it out (assuming \(n! \neq 0\)):
\[ \frac{1}{(n-5)!} = 20 \cdot \frac{1}{(n-3)!} \]

### Step 4: Rewrite the Factorials
Rearranging gives us:
\[ (n-3)(n-4) = 20 \]
This is because \((n-3)! = (n-3)(n-4)(n-5)!\).

### Step 5: Expand and Rearrange
Expanding the left side:
\[ n^2 - 7n + 12 = 20 \]
Now, rearranging gives:
\[ n^2 - 7n - 8 = 0 \]

### Step 6: Factor the Quadratic Equation
Next, we can factor the quadratic equation:
\[ (n - 8)(n + 1) = 0 \]

### Step 7: Solve for \(n\)
Setting each factor to zero gives us:
\[ n - 8 = 0 \quad \text{or} \quad n + 1 = 0 \]
Thus, we find:
\[ n = 8 \quad \text{or} \quad n = -1 \]

### Step 8: Determine Validity of Solutions
Since \(n\) must be a non-negative integer (as it represents the total number of items in permutations), we discard \(n = -1\). Therefore, the only valid solution is:
\[ n = 8 \]

### Final Answer
The value of \(n\) is \(8\).

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