Given the equation:
\[ \frac{22 P_{r+1}}{20 P_{r+2}} = \frac{11}{52}. \]Applying the permutation formula \(^{n}P_r = \frac{n!}{(n - r)!}\), we expand the expression:
\[ \frac{\frac{22!}{(22 - (r+1))!}}{\frac{20!}{(20 - (r+2))!}} = \frac{11}{52} \]This simplifies to:
\[ \frac{22!}{(21 - r)!} \times \frac{(18 - r)!}{20!} = \frac{11}{52} \]After canceling common factorial terms:
\[ \frac{22 \times 21 \times 20!}{(21 - r)!} \times \frac{(18 - r)!}{20!} = \frac{11}{52} \]Further simplification yields:
\[ \frac{22 \times 21}{(21 - r)(20 - r)} = \frac{11}{52} \]Cross-multiplying gives:
\[ 22 \times 21 \times 52 = 11 \times (21 - r) \times (20 - r) \] \[ 22 \times 21 \times 52 = 11 \times (21 - r) \times (20 - r) \]Dividing both sides by 11:
\[ 2 \times 21 \times 52 = (21 - r)(20 - r) \]To solve for \(r\):
\[ (21 - r)(20 - r) = 14 \times 13 \]This implies:
\[ 21 - r = 14, \quad 20 - r = 13 \]Which results in:
\[ r = 7 \]The calculated value of \( r \) is 7.