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, we are given two equations involving combinations and permutations:

1. \( \binom{n}{x} = 56 \)
2. \( P(n, x) = 336 \)

We need to find the value of \( n \).

### Step-by-Step Solution:

**Step 1: Write the formulas for combinations and permutations.**

The combination formula is given by:
\[ \binom{n}{x} = \frac{n!}{x!(n-x)!} \]
The permutation formula is given by:
\[ P(n, x) = \frac{n!}{(n-x)!} \]

**Step 2: Set up the equations.**

From the given information, we can set up the following equations:
\[ \frac{n!}{x!(n-x)!} = 56 \quad \text{(Equation 1)} \]
\[ \frac{n!}{(n-x)!} = 336 \quad \text{(Equation 2)} \]

**Step 3: Express \( n! \) from both equations.**

From Equation 1, we can express \( n! \) as:
\[ n! = 56 \cdot x!(n-x)! \]

From Equation 2, we can express \( n! \) as:
\[ n! = 336 \cdot (n-x)! \]

**Step 4: Set the two expressions for \( n! \) equal to each other.**

Equating the two expressions for \( n! \):
\[ 56 \cdot x!(n-x)! = 336 \cdot (n-x)! \]

**Step 5: Simplify the equation.**

We can cancel \( (n-x)! \) from both sides (assuming \( n \neq x \)):
\[ 56 \cdot x! = 336 \]

**Step 6: Solve for \( x! \).**

Dividing both sides by 56 gives:
\[ x! = \frac{336}{56} = 6 \]

**Step 7: Determine the value of \( x \).**

The factorial \( x! = 6 \) corresponds to:
\[ x = 3 \quad \text{(since \( 3! = 6 \))} \]

**Step 8: Substitute \( x \) back into Equation 2.**

Now we substitute \( x = 3 \) into Equation 2:
\[ \frac{n!}{(n-3)!} = 336 \]

**Step 9: Simplify the equation.**

This simplifies to:
\[ n(n-1)(n-2) = 336 \]

**Step 10: Factor 336.**

We can factor 336 as:
\[ 336 = 8 \times 7 \times 6 \]

**Step 11: Set up the equation.**

Thus, we have:
\[ n(n-1)(n-2) = 8 \times 7 \times 6 \]

**Step 12: Solve for \( n \).**

By inspection, if \( n = 8 \):
\[ 8 \times 7 \times 6 = 336 \]

Thus, \( n = 8 \).

### Final Answer:
The value of \( n \) is \( \boxed{8} \).