Question

If \( 22 P_{r+1} : 20 P_{r+2} = 11 : 52 \), then \( r \) is equal to:

• A. 3
• B. 5
• C. 7
• D. 9

Hint:

For problems involving permutations, simplify the expressions and solve step-by-step to isolate the variable.

Answer 1

The Correct Option is C

We are given:

\[ \frac{22 P_{r+1}}{20 P_{r+2}} = \frac{11}{52}. \]

Using the formula for permutations:

\[ ^{n}P_r = \frac{n!}{(n - r)!} \]

Expanding the given expression:

\[ \frac{\frac{22!}{(22 - (r+1))!}}{\frac{20!}{(20 - (r+2))!}} = \frac{11}{52} \]

Simplifying:

\[ \frac{22!}{(21 - r)!} \times \frac{(18 - r)!}{20!} = \frac{11}{52} \]

Canceling common factorial terms:

\[ \frac{22 \times 21 \times 20!}{(21 - r)!} \times \frac{(18 - r)!}{20!} = \frac{11}{52} \]

Simplifying further:

\[ \frac{22 \times 21}{(21 - r)(20 - r)} = \frac{11}{52} \]

Cross multiplying:

\[ 22 \times 21 \times 52 = 11 \times (21 - r) \times (20 - r) \] \[ 22 \times 21 \times 52 = 11 \times (21 - r) \times (20 - r) \]

Dividing both sides by 11:

\[ 2 \times 21 \times 52 = (21 - r)(20 - r) \]

Solving for \(r\):

\[ (21 - r)(20 - r) = 14 \times 13 \] \[ 21 - r = 14, \quad 20 - r = 13 \] \[ r = 7 \]

Thus, the value of \( r \) is 7, which matches option (C).