Question

If \( ^{11}P_r = 7920 \), then the value of \( r \) is equal to

• A. \(7 \)
• B. \(6 \)
• C. \(5 \)
• D. \(4 \)
• E. \(3 \)

Hint:

For permutation equations, try small values of \(r\) instead of expanding factorials fully.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The notation \({}^{n}P_r\) represents the number of permutations of `n` items taken `r` at a time. It calculates the number of ways to arrange `r` items selected from a set of `n` distinct items.
Step 2: Key Formula or Approach:
The formula for permutations is: \[ {}^nP_r = \frac{n!}{(n-r)!} = n \times (n-1) \times (n-2) \times \dots \times (n-r+1) \] We are given \({}^{11}P_r = 7920\). The most practical way to solve this is to start multiplying numbers downwards from 11 and see how many terms are needed to reach 7920.
Step 3: Detailed Explanation:
We need to find `r` such that the product of `r` consecutive integers starting from 11 is 7920. Let's calculate the product step-by-step:
For r = 1: \({}^{11}P_1 = 11\)
For r = 2: \({}^{11}P_2 = 11 \times 10 = 110\)
For r = 3: \({}^{11}P_3 = 11 \times 10 \times 9 = 110 \times 9 = 990\)
For r = 4: \({}^{11}P_4 = 11 \times 10 \times 9 \times 8 = 990 \times 8 = 7920\)
We have reached the target value of 7920 after multiplying 4 terms.
Step 4: Final Answer:
The value of r is 4.