if ` .^(m+n) P_(2)=56 and .^(m-n)P_(3)=24`, then `(.^(m)P_(3))/(.^(n)P_(2))` equals

Options

A. 20
B. 40
C. 60
D. 80

Answer

To solve the problem, we need to find the value of \(\frac{mP_3}{nP_2}\) given the equations \(^{(m+n)}P_2 = 56\) and \(^{(m-n)}P_3 = 24\).

### Step-by-Step Solution:

1. **Use the formula for permutations**:
The formula for permutations is given by:
\[ ^nP_r = \frac{n!}{(n-r)!} \]
Therefore, we can rewrite the first equation:
\[ ^{(m+n)}P_2 = \frac{(m+n)!}{(m+n-2)!} = 56 \]

2. **Simplify the first equation**:
This can be simplified to:
\[ (m+n)(m+n-1) = 56 \]
We can factor 56 as \(8 \times 7\). Thus, we have:
\[ m+n = 8 \quad \text{and} \quad m+n-1 = 7 \]
This gives us our first equation:
\[ m+n = 8 \quad \text{(Equation 1)} \]

3. **Use the second equation**:
Now, we apply the same permutation formula to the second equation:
\[ ^{(m-n)}P_3 = \frac{(m-n)!}{(m-n-3)!} = 24 \]

4. **Simplify the second equation**:
This can be simplified to:
\[ (m-n)(m-n-1)(m-n-2) = 24 \]
We can factor 24 as \(4 \times 3 \times 2\). Thus, we have:
\[ m-n = 4 \quad \text{(Equation 2)} \]

5. **Solve the system of equations**:
Now we have two equations:
- \(m+n = 8\) (Equation 1)
- \(m-n = 4\) (Equation 2)

We can solve these equations simultaneously. Adding both equations:
\[ (m+n) + (m-n) = 8 + 4 \]
This simplifies to:
\[ 2m = 12 \implies m = 6 \]

6. **Find \(n\)**:
Substitute \(m = 6\) back into Equation 1:
\[ 6 + n = 8 \implies n = 2 \]

7. **Calculate \(mP_3\) and \(nP_2\)**:
Now we need to calculate \(\frac{mP_3}{nP_2}\):
\[ mP_3 = ^6P_3 = \frac{6!}{(6-3)!} = \frac{6!}{3!} = \frac{6 \times 5 \times 4}{1} = 120 \]
\[ nP_2 = ^2P_2 = \frac{2!}{(2-2)!} = 2! = 2 \]

8. **Final Calculation**:
Now we can find:
\[ \frac{mP_3}{nP_2} = \frac{120}{2} = 60 \]

Thus, the final answer is:
\[ \boxed{60} \]