If ` ""^n C_x=56 and ""^(n)P_x = 336 ` find n and x .

Options

A. 7,3,
B. 8,4
C. 8,3
D. 9,6

Answer

To solve the problem, we need to find the values of \( n \) and \( x \) given the equations:

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

### Step 1: Write the equations in terms of factorials

Using the definitions of combinations and permutations, we can express these equations as follows:

\[ \binom{n}{x} = \frac{n!}{x!(n-x)!} = 56 \]

\[ P(n, x) = \frac{n!}{(n-x)!} = 336 \]

### Step 2: Divide the equations

We can divide the first equation by the second equation to eliminate \( n! \):

\[ \frac{\frac{n!}{x!(n-x)!}}{\frac{n!}{(n-x)!}} = \frac{56}{336} \]

This simplifies to:

\[ \frac{1}{x!} = \frac{56}{336} \]

### Step 3: Simplify the right side

Calculating \( \frac{56}{336} \):

\[ \frac{56}{336} = \frac{1}{6} \]

Thus, we have:

\[ \frac{1}{x!} = \frac{1}{6} \]

### Step 4: Solve for \( x \)

Taking the reciprocal gives us:

\[ x! = 6 \]

The factorial of \( x \) equals 6, which means:

\[ x = 3 \]

### Step 5: Substitute \( x \) back to find \( n \)

Now that we have \( x = 3 \), we can substitute it back into the permutation equation:

\[ P(n, 3) = 336 \]

This means:

\[ \frac{n!}{(n-3)!} = 336 \]

### Step 6: Expand the permutation equation

Expanding this gives:

\[ n(n-1)(n-2) = 336 \]

### Step 7: Solve for \( n \)

Now we need to find \( n \) such that:

\[ n(n-1)(n-2) = 336 \]

We can try different values of \( n \):

- For \( n = 7 \):
\[ 7 \times 6 \times 5 = 210 \quad (\text{too low}) \]

- For \( n = 8 \):
\[ 8 \times 7 \times 6 = 336 \quad (\text{correct}) \]

Thus, \( n = 8 \).

### Final Answer

The values of \( n \) and \( x \) are:

\[ n = 8, \quad x = 3 \]