Question

If \( ^{56}P_{r+6} : \, ^{54}P_{r+3} = 30800 : 1 \), then \( r \) is equal to

• A. \( 69 \)
• B. \( 41 \)
• C. \( 51 \)
• D. \( 61 \)
• E. \( 49 \)

Hint:

In permutation ratios, always expand factorials to cancel terms instead of calculating full factorial values.

Answer 1

The Correct Option is B

Concept: Use permutation formula: \[ ^nP_r = \frac{n!}{(n-r)!} \] and simplify ratios carefully by expanding factorials.

Step 1: Write both permutations using factorial form: \[ ^{56}P_{r+6} = \frac{56!}{(56-(r+6))!} = \frac{56!}{(50-r)!} \] \[ ^{54}P_{r+3} = \frac{54!}{(54-(r+3))!} = \frac{54!}{(51-r)!} \]

Step 2: Form the ratio: \[ \frac{^{56}P_{r+6}}{^{54}P_{r+3}} = \frac{56!}{(50-r)!} \cdot \frac{(51-r)!}{54!} \]

Step 3: Simplify factorials step-by-step: \[ \frac{56!}{54!} = 56 \times 55 \] and \[ \frac{(51-r)!}{(50-r)!} = (51-r) \] So the ratio becomes: \[ 56 \times 55 \times (51-r) \]

Step 4: Use given condition: \[ 56 \cdot 55 \cdot (51-r) = 30800 \] First compute: \[ 56 \cdot 55 = 3080 \] So: \[ 3080(51-r)=30800 \]

Step 5: Solve equation: \[ 51-r = \frac{30800}{3080} = 10 \] \[ r = 51 - 10 = 41 \]

Step 6: Verify logically: Since \( r+6 \le 56 \Rightarrow r \le 50 \), value \( r=41 \) is valid.