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).

Answer 2

Step 1: Write down the given ratio
\(\frac{{}^{22}P_{r+1}}{\,{}^{20}P_{r+2}} = \frac{11}{52}\)

Step 2: Recall the permutation formula
\({}^nP_k = \frac{n!}{(n-k)!}\)

Step 3: Express each permutation term
\({}^{22}P_{r+1} = \frac{22!}{(22 - (r+1))!} = \frac{22!}{(21 - r)!}\)
\({}^{20}P_{r+2} = \frac{20!}{(20 - (r+2))!} = \frac{20!}{(18 - r)!}\)

Step 4: Substitute into the ratio
\[ \frac{\frac{22!}{(21 - r)!}}{\frac{20!}{(18 - r)!}} = \frac{11}{52} \quad \Rightarrow \quad \frac{22!}{(21 - r)!} \times \frac{(18 - r)!}{20!} = \frac{11}{52} \]

Step 5: Simplify factorial expressions
\[ \frac{22 \times 21 \times 20!}{(21 - r)!} \times \frac{(18 - r)!}{20!} = \frac{11}{52} \] Cancel \(20!\):
\[ \frac{22 \times 21 \times (18 - r)!}{(21 - r)!} = \frac{11}{52} \]

Step 6: Express \((21 - r)!\) in terms of \((18 - r)!\)
\[ (21 - r)! = (21 - r)(20 - r)(19 - r)(18 - r)! \] So substitute:
\[ \frac{22 \times 21 \times (18 - r)!}{(21 - r)(20 - r)(19 - r)(18 - r)!} = \frac{11}{52} \] Cancel \((18 - r)!\):
\[ \frac{22 \times 21}{(21 - r)(20 - r)(19 - r)} = \frac{11}{52} \]

Step 7: Simplify numerator
\(22 \times 21 = 462\)
So:
\[ \frac{462}{(21 - r)(20 - r)(19 - r)} = \frac{11}{52} \] Cross multiply:
\[ 462 \times 52 = 11 \times (21 - r)(20 - r)(19 - r) \]

Step 8: Calculate left side
\(462 \times 52 = 24024\)
So:
\[ 24024 = 11 \times (21 - r)(20 - r)(19 - r) \] Divide both sides by 11:
\[ (21 - r)(20 - r)(19 - r) = \frac{24024}{11} = 2184 \]

Step 9: Find integer \(r\) such that product equals 2184
Try \(r = 7\):
\[ 21 - 7 = 14, \quad 20 - 7 = 13, \quad 19 - 7 = 12 \] Calculate product:
\(14 \times 13 \times 12 = 2184\)

Step 10: Conclusion
Since the product matches 2184 at \(r = 7\), the value of \(r\) is:
7