Question:
If \( ^nP_4 = 5(^nP_3) \), then the value of \( n \) is equal to:
If \( ^nP_4 = 5(^nP_3) \), then the value of \( n \) is equal to:
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A useful identity to remember is \( ^nP_r = (n-r+1) \cdot ^nP_{r-1} \). In this problem, \( ^nP_4 = (n-4+1) \cdot ^nP_3 = (n-3) \cdot ^nP_3 \). Setting this equal to \( 5 \cdot ^nP_3 \) gives \( n-3=5 \) instantly.
Updated On: May 2, 2026
- \( 5 \)
- \( 6 \)
- \( 7 \)
- \( 8 \)
- \( 9 \)
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The Correct Option is D
Solution and Explanation
Concept:
The formula for permutations of \( n \) objects taken \( r \) at a time is given by:
\[
^nP_r = \frac{n!}{(n-r)!} = n(n-1)(n-2)\cdots(n-r+1)
\]
This represents the number of ways to arrange a subset of items where order matters.
Step 1: Expand both sides of the equation using the permutation formula.
For the left side: \[ ^nP_4 = n(n-1)(n-2)(n-3) \] For the right side: \[ 5(^nP_3) = 5 \times n(n-1)(n-2) \]
Step 2: Set up the equation and simplify.
\[ n(n-1)(n-2)(n-3) = 5n(n-1)(n-2) \] Since \( n \) must be at least 4 (from \( ^nP_4 \)), we know that \( n \), \( n-1 \), and \( n-2 \) are non-zero. We can safely divide both sides by \( n(n-1)(n-2) \): \[ n - 3 = 5 \]
Step 3: Solve for \( n \).
\[ n = 5 + 3 \] \[ n = 8 \]
Step 1: Expand both sides of the equation using the permutation formula.
For the left side: \[ ^nP_4 = n(n-1)(n-2)(n-3) \] For the right side: \[ 5(^nP_3) = 5 \times n(n-1)(n-2) \]
Step 2: Set up the equation and simplify.
\[ n(n-1)(n-2)(n-3) = 5n(n-1)(n-2) \] Since \( n \) must be at least 4 (from \( ^nP_4 \)), we know that \( n \), \( n-1 \), and \( n-2 \) are non-zero. We can safely divide both sides by \( n(n-1)(n-2) \): \[ n - 3 = 5 \]
Step 3: Solve for \( n \).
\[ n = 5 + 3 \] \[ n = 8 \]
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