Find 'n' if `2 ""^(5)P_(3)= ""^(n)P_(4)`

Answer

To solve the equation \( 2 \cdot {}^{5}P_{3} = {}^{n}P_{4} \), we will follow these steps:

### Step 1: Expand \( {}^{5}P_{3} \)
The formula for permutations is given by:
\[ {}^{r}P_{n} = \frac{n!}{(n-r)!} \]
Thus, we can write:
\[ {}^{5}P_{3} = \frac{5!}{(5-3)!} = \frac{5!}{2!} \]

### Step 2: Calculate \( 5! \) and \( 2! \)
Calculating the factorials:
\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]
\[ 2! = 2 \times 1 = 2 \]
Now substituting these values back:
\[ {}^{5}P_{3} = \frac{120}{2} = 60 \]

### Step 3: Substitute back into the equation
Now substituting this value into the original equation:
\[ 2 \cdot 60 = {}^{n}P_{4} \]
This simplifies to:
\[ 120 = {}^{n}P_{4} \]

### Step 4: Expand \( {}^{n}P_{4} \)
Using the permutations formula again:
\[ {}^{n}P_{4} = \frac{n!}{(n-4)!} \]
So we have:
\[ 120 = \frac{n!}{(n-4)!} \]

### Step 5: Simplify the equation
This can be rewritten as:
\[ 120 = n \cdot (n-1) \cdot (n-2) \cdot (n-3) \]
This means we need to find \( n \) such that:
\[ n(n-1)(n-2)(n-3) = 120 \]

### Step 6: Solve for \( n \)
We can test integer values for \( n \):
- For \( n = 5 \):
\[ 5 \cdot 4 \cdot 3 \cdot 2 = 120 \quad \text{(This works)} \]

Thus, the value of \( n \) is:
\[ \boxed{5} \]