Question

If \( {}^{27}P_{r+7} = 7722 \cdot {}^{25}P_{r+4} \), then r =

• A. \( 9 \)
• B. \( 12 \)
• C. \( 11 \)
• D. \( 10 \) Correct Answer

Hint:

Recall the definition of permutation: \( {}^n P_k = \frac{n!}{(n-k)!} \). Ensure that \( n \ge k \) and \( k \ge 0 \). When simplifying ratios of factorials, \( \frac{A!}{(A-B)!} = A(A-1)\dots(A-B+1) \). Also, \( (N)! = N \cdot (N-1)! \).

Answer 1

The Correct Option is D

Step 1: Use the formula for permutations \( {}^n P_k = \frac{n!}{(n-k)!} \).
The given equation is \( {}^{27}P_{r+7} = 7722 \cdot {}^{25}P_{r+4} \).
LHS: \( {}^{27}P_{r+7} = \frac{27!}{(27-(r+7))!} = \frac{27!}{(20-r)!} \).
RHS: \( {}^{25}P_{r+4} = \frac{25!}{(25-(r+4))!} = \frac{25!}{(21-r)!} \).

Step 2: Substitute these into the equation.
\[ \frac{27!}{(20-r)!} = 7722 \cdot \frac{25!}{(21-r)!} \]
Step 3: Simplify the factorial expressions.
\( \frac{27 \cdot 26 \cdot 25!}{(20-r)!} = 7722 \cdot \frac{25!}{(21-r) \cdot (20-r)!} \) Assuming \( (20-r)! \ne 0 \) and \( 25! \ne 0 \), we can cancel them: \[ 27 \cdot 26 = \frac{7722}{21-r} \]
Step 4: Solve for \(r\).
\[ 702 = \frac{7722}{21-r} \] \[ 21-r = \frac{7722}{702} \] Calculate the division: \( 7722 \div 702 \).
\( 702 \times 10 = 7020 \).
\( 7722 - 7020 = 702 \).
So, \( 7722 = 702 \times 10 + 702 = 702 \times 11 \).
Thus, \( \frac{7722}{702} = 11 \).
\[ 21-r = 11 \] \[ r = 21 - 11 \] \[ r = 10 \]
Step 5: Verify the conditions for permutations.
For \( {}^{27}P_{r+7} \): \( r+7 \ge 0 \Rightarrow r \ge -7 \).
Also, \( 27 \ge r+7 \Rightarrow r \le 20 \).
For \( {}^{25}P_{r+4} \): \( r+4 \ge 0 \Rightarrow r \ge -4 \).
Also, \( 25 \ge r+4 \Rightarrow r \le 21 \).
Also, from denominators of factorials: \( 20-r \ge 0 \Rightarrow r \le 20 \) and \( 21-r \ge 0 \Rightarrow r \le 21 \).
The value \( r=10 \) satisfies all these conditions: \( 10 \ge -7 \), \( 10 \le 20 \), \( 10 \ge -4 \), \( 10 \le 21 \).
The values selected are \(r+7=17\) and \(r+4=14\), which are valid.
This matches option (4).