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: Write the permutation formula.
We use $\,^nP_k = \dfrac{n!}{(n-k)!}$. The condition $\,^nP_r = \,^nP_s$ becomes
\[ \frac{n!}{(n-r)!} = \frac{n!}{(n-s)!} \]

Step 2: Cancel and compare.
Drop the common $n!$ from both sides.
\[ (n-r)! = (n-s)! \]

Step 3: Use the special factorial pair.
Two different whole numbers give equal factorials only for the pair $0! = 1! = 1$. Since $r < s$, we have $n - r > n - s$, so the bigger one is $1$ and the smaller is $0$.
\[ n - r = 1, \qquad n - s = 0 \]
This gives $r = n - 1$ and $s = n$, which fits $0 < r < s \le n$.

Step 4: Find $r - s$.
\[ r - s = (n-1) - n = -1 \]
That is option 1.
\[ \boxed{-1} \]