Question

If \( 0 < r < s \le n \) and \( ^nP_r = ^nP_s \), then the value of \( (r - s) \) is :

• A. \( -1 \)
• B. \( -2n - 1 \)
• C. \( -2 \)
• D. \( -2n - 2 \) Correct Answer: (A) \( -1 \) Solution:

Hint:

Remember the "Last Two Rule" for permutations:
\( ^nP_n = ^nP_{n-1} = n! \).
For combinations, the rule is symmetric: \( ^nC_r = ^nC_{n-r} \).
Don't get confused between the two; permutations only have this single specific collision of values.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The notation \( ^nP_k \) represents the number of permutations of \( n \) distinct objects taken \( k \) at a time. The formula is defined as \( \frac{n!}{(n-k)!} \).
Usually, as you increase the number of items you pick (\( k \)), the number of ways to arrange them increases significantly because each new slot adds more possibilities. However, factorials have a unique property at the very end of their range: \( 1! \) and \( 0! \) are both equal to 1. This mathematical convention (which arises from the definition of the Gamma function and the identity for the empty product) implies that the number of ways to arrange almost all items (\( n-1 \)) is the same as the number of ways to arrange every single item (\( n \)).

Step 2: Key Formula or Approach:
The general formula is \( ^nP_k = \frac{n!}{(n-k)!} \).
We are given \( ^nP_r = ^nP_s \) with \( r < s \).
This implies \( \frac{n!}{(n-r)!} = \frac{n!}{(n-s)!} \).
Since \( n! \) is a non-zero constant for \( n \ge 1 \), we can divide both sides by \( n! \):
\[ \frac{1}{(n-r)!} = \frac{1}{(n-s)!} \implies (n-r)! = (n-s)! \]
Since \( x! = y! \) for distinct non-negative integers \( x, y \) only when one is 0 and the other is 1, we look for that specific scenario.

Step 3: Detailed Explanation:
Step 3.1: Identifying the values for \( r \) and \( s \).
We know that \( 0! = 1 \) and \( 1! = 1 \).
For the equation \( (n-r)! = (n-s)! \) to hold for distinct values of \( r \) and \( s \), the only possibility within the factorial domain is:
One side equals \( 0! \) and the other side equals \( 1! \).
Since we are given \( r < s \), then \( n-r \) must be larger than \( n-s \).
Thus:
\[ n - r = 1 \]
\[ n - s = 0 \]
Step 3.2: Solving for \( r \) and \( s \).
From \( n - s = 0 \), we get \( s = n \).
From \( n - r = 1 \), we get \( r = n - 1 \).
Checking constraints: \( 0 < n-1 < n \le n \). This fits the given condition \( 0 < r < s \le n \).
Step 3.3: Calculating \( r - s \).
\[ r - s = (n - 1) - n \]
\[ r - s = n - 1 - n = -1 \]
This matches Option (A).

Step 4: Final Answer:
The only case where \( ^nP_r = ^nP_s \) for \( r \neq s \) is when we compare arranging all \( n \) items versus \( n-1 \) items. In both cases, the result is \( n! \). Calculating the difference between these indices (\( n-1 \) and \( n \)) gives \( -1 \). Thus, Option (A) is the correct answer.