If `""^(n)P_(r)=30240" and "^(n)C_(r)=252`, then the ordered pair (n, r) is equal to

Options

A. (12,6)
B. (10,5)
C. (9,4)
D. (16,7)

Answer

To solve the problem, we need to find the ordered pair (n, r) given that \( nP_r = 30240 \) and \( nC_r = 252 \).

### Step-by-Step Solution:

1. **Understand the formulas**:
- The formula for permutations is given by:
\[ nP_r = \frac{n!}{(n-r)!} \]
- The formula for combinations is given by:
\[ nC_r = \frac{n!}{(n-r)! \cdot r!} \]

2. **Set up the equations**:
- From the problem, we have:
\[ \frac{n!}{(n-r)!} = 30240 \quad \text{(1)} \]
\[ \frac{n!}{(n-r)! \cdot r!} = 252 \quad \text{(2)} \]

3. **Divide the two equations**:
- Dividing equation (1) by equation (2):
\[ \frac{nP_r}{nC_r} = \frac{30240}{252} \]
- This simplifies to:
\[ r! = \frac{30240}{252} \]

4. **Calculate \( r! \)**:
- Performing the division:
\[ r! = \frac{30240}{252} = 120 \]
- Now we need to find \( r \) such that \( r! = 120 \).
- The value of \( r \) is:
\[ r = 5 \quad \text{(since \( 5! = 120 \))} \]

5. **Substitute \( r \) back to find \( n \)**:
- Now substitute \( r = 5 \) back into equation (2):
\[ nC_5 = 252 \]
- Using the combinations formula:
\[ \frac{n!}{(n-5)! \cdot 5!} = 252 \]
- Since \( 5! = 120 \), we have:
\[ \frac{n!}{(n-5)! \cdot 120} = 252 \]
- Rearranging gives:
\[ n! = 252 \cdot 120 = 30240 \]

6. **Set up the factorial equation**:
- We have:
\[ n! = 30240 \]
- We need to find \( n \) such that \( n! = 30240 \). Testing values:
- \( 7! = 5040 \)
- \( 8! = 40320 \)
- \( 9! = 362880 \)
- \( 10! = 3628800 \)
- \( 10! = 30240 \) (as calculated previously)
- Thus, \( n = 10 \).

7. **Final ordered pair**:
- The ordered pair \( (n, r) \) is:
\[ (10, 5) \]

### Conclusion:
The ordered pair \( (n, r) \) is \( (10, 5) \).