Find n when `""^(n-1)P_(4): ""^(n)P_(5)=1:9`, then n = …………….

Answer

To solve the problem, we need to find the value of \( n \) such that

\[ \frac{{(n-1)P_4}}{{nP_5}} = \frac{1}{9} \]

### Step 1: Write the formula for permutations
The formula for permutations is given by:

\[ nPr = \frac{n!}{(n-r)!} \]

### Step 2: Apply the formula to the given problem
Using the formula, we can express \( (n-1)P_4 \) and \( nP_5 \):

\[ (n-1)P_4 = \frac{(n-1)!}{(n-1-4)!} = \frac{(n-1)!}{(n-5)!} \]

\[ nP_5 = \frac{n!}{(n-5)!} \]

### Step 3: Substitute these into the ratio
Now substituting these into the ratio gives us:

\[ \frac{{(n-1)!/(n-5)!}}{{n!/(n-5)!}} = \frac{(n-1)!}{n!} \]

### Step 4: Simplify the expression
We can simplify \( \frac{(n-1)!}{n!} \):

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

### Step 5: Set up the equation
Now we can set up the equation based on the original problem statement:

\[ \frac{1}{n} = \frac{1}{9} \]

### Step 6: Solve for \( n \)
Cross-multiplying gives:

\[ 1 \cdot 9 = 1 \cdot n \implies n = 9 \]

Thus, the value of \( n \) is:

\[ \boxed{9} \]