If `""^(56)P_(r+ 6)`:`""^(54)P_(r+3)` = 30800, find `""^(r)P_(2)`.

Options

A. 1840
B. 2640
C. 1640
D. 820

Answer

To solve the problem, we need to find the value of \( rP2 \) given the equation:

\[ \frac{^{56}P_{r+6}}{^{54}P_{r+3}} = 30800 \]

### Step-by-step Solution:

1. **Write the permutations in terms of factorials:**
The formula for permutations is given by:
\[ ^nP_r = \frac{n!}{(n-r)!} \]
Therefore, we can express \( ^{56}P_{r+6} \) and \( ^{54}P_{r+3} \) as follows:
\[ ^{56}P_{r+6} = \frac{56!}{(56 - (r + 6))!} = \frac{56!}{(50 - r)!} \]
\[ ^{54}P_{r+3} = \frac{54!}{(54 - (r + 3))!} = \frac{54!}{(51 - r)!} \]

2. **Set up the equation:**
Substituting these expressions into the original equation gives:
\[ \frac{\frac{56!}{(50 - r)!}}{\frac{54!}{(51 - r)!}} = 30800 \]
This simplifies to:
\[ \frac{56! \cdot (51 - r)!}{54! \cdot (50 - r)!} = 30800 \]

3. **Simplify the left-hand side:**
We can further simplify:
\[ \frac{56 \cdot 55 \cdot 54! \cdot (51 - r)!}{54! \cdot (50 - r)!} = \frac{56 \cdot 55 \cdot (51 - r)!}{(50 - r)!} \]
This reduces to:
\[ 56 \cdot 55 \cdot \frac{(51 - r)!}{(50 - r)!} = 56 \cdot 55 \cdot (51 - r) \]

4. **Set the equation equal to 30800:**
Now we have:
\[ 56 \cdot 55 \cdot (51 - r) = 30800 \]

5. **Calculate \( 56 \cdot 55 \):**
First, calculate \( 56 \cdot 55 \):
\[ 56 \cdot 55 = 3080 \]
Therefore, the equation becomes:
\[ 3080 \cdot (51 - r) = 30800 \]

6. **Solve for \( 51 - r \):**
Dividing both sides by 3080 gives:
\[ 51 - r = 10 \]
Thus, we find:
\[ r = 41 \]

7. **Find \( ^{r}P_{2} \):**
Now we need to calculate \( ^{41}P_{2} \):
\[ ^{41}P_{2} = \frac{41!}{(41 - 2)!} = \frac{41!}{39!} = 41 \cdot 40 \]

8. **Calculate the final result:**
\[ 41 \cdot 40 = 1640 \]

### Final Answer:
The value of \( ^{r}P_{2} \) is \( 1640 \).