If `""^(n)C_(x) = 56` and `""^(n)P_(x) = 336 `, then n is equal to

Options

A. 7
B. 8
C. 6
D. 10

Answer

To solve the problem where \( \binom{n}{x} = 56 \) and \( P(n, x) = 336 \), we need to find the value of \( n \).

### Step-by-Step Solution:

1. **Understanding the relationships**:
We know that:
\[ P(n, x) = \frac{n!}{(n-x)!} \]
and
\[ \binom{n}{x} = \frac{n!}{x!(n-x)!} \]

2. **Setting up the equations**:
From the given information:
\[ \binom{n}{x} = 56 \implies \frac{n!}{x!(n-x)!} = 56 \]
\[ P(n, x) = 336 \implies \frac{n!}{(n-x)!} = 336 \]

3. **Expressing \( n! \)**:
From the second equation, we can express \( n! \):
\[ n! = 336 \cdot (n-x)! \]

4. **Substituting \( n! \) into the first equation**:
Substitute \( n! \) from the above equation into the first equation:
\[ \frac{336 \cdot (n-x)!}{x!(n-x)!} = 56 \]
Simplifying this gives:
\[ \frac{336}{x!} = 56 \]

5. **Solving for \( x! \)**:
Rearranging the equation:
\[ x! = \frac{336}{56} = 6 \]

6. **Finding \( x \)**:
Since \( 3! = 6 \), we have:
\[ x = 3 \]

7. **Using \( x \) to find \( n \)**:
Now substitute \( x = 3 \) back into the equation for \( P(n, x) \):
\[ P(n, 3) = 336 \implies \frac{n!}{(n-3)!} = 336 \]
This simplifies to:
\[ n(n-1)(n-2) = 336 \]

8. **Setting up the cubic equation**:
Rearranging gives us:
\[ n^3 - 3n^2 + 2n - 336 = 0 \]

9. **Finding the roots of the cubic equation**:
We can test integer values for \( n \). Let's try \( n = 8 \):
\[ 8^3 - 3(8^2) + 2(8) - 336 = 512 - 192 + 16 - 336 = 0 \]
Thus, \( n = 8 \) satisfies the equation.

10. **Conclusion**:
Therefore, the value of \( n \) is:
\[ \boxed{8} \]