If `""^(n+2) C_8: "^(n-2) P_4=57:16` then n=

Options

A. 20
B. 19
C. 18
D. 17

Answer

To solve the equation given in the question, we have:

\[ \frac{{^{(n+2)}C_8}}{{^{(n-2)}P_4}} = \frac{57}{16} \]

### Step 1: Write the formulas for combinations and permutations

The formula for combinations is given by:

\[ ^{n}C_r = \frac{n!}{r!(n-r)!} \]

And the formula for permutations is given by:

\[ ^{n}P_r = \frac{n!}{(n-r)!} \]

### Step 2: Substitute the formulas into the equation

Using the formulas, we can rewrite the left-hand side of the equation:

\[ ^{(n+2)}C_8 = \frac{(n+2)!}{8!(n+2-8)!} = \frac{(n+2)!}{8!(n-6)!} \]

And for the permutation:

\[ ^{(n-2)}P_4 = \frac{(n-2)!}{(n-2-4)!} = \frac{(n-2)!}{(n-6)!} \]

### Step 3: Substitute these into the equation

Now we can substitute these into our original equation:

\[ \frac{\frac{(n+2)!}{8!(n-6)!}}{\frac{(n-2)!}{(n-6)!}} = \frac{57}{16} \]

### Step 4: Simplify the equation

The \( (n-6)! \) cancels out:

\[ \frac{(n+2)!}{8! (n-2)!} = \frac{57}{16} \]

### Step 5: Rewrite \( (n+2)! \)

We can express \( (n+2)! \) in terms of \( (n-2)! \):

\[ (n+2)! = (n+2)(n+1)(n)(n-1)(n-2)! \]

Substituting this back gives:

\[ \frac{(n+2)(n+1)n(n-1)(n-2)!}{8!(n-2)!} = \frac{57}{16} \]

### Step 6: Cancel \( (n-2)! \)

Cancelling \( (n-2)! \):

\[ \frac{(n+2)(n+1)n(n-1)}{8!} = \frac{57}{16} \]

### Step 7: Multiply both sides by \( 8! \)

Now, multiply both sides by \( 8! \):

\[ (n+2)(n+1)n(n-1) = 57 \cdot \frac{8!}{16} \]

### Step 8: Calculate \( 8! \)

Calculating \( 8! \):

\[ 8! = 40320 \]

So,

\[ \frac{8!}{16} = \frac{40320}{16} = 2520 \]

### Step 9: Substitute back into the equation

Now we have:

\[ (n+2)(n+1)n(n-1) = 57 \cdot 2520 \]

Calculating \( 57 \cdot 2520 \):

\[ 57 \cdot 2520 = 143640 \]

### Step 10: Set up the equation

Now we have:

\[ (n+2)(n+1)n(n-1) = 143640 \]

### Step 11: Solve for \( n \)

To find \( n \), we can try integer values. Testing \( n = 19 \):

\[ (19+2)(19+1)(19)(19-1) = 21 \cdot 20 \cdot 19 \cdot 18 \]

Calculating:

\[ 21 \cdot 20 = 420 \]
\[ 420 \cdot 19 = 7980 \]
\[ 7980 \cdot 18 = 143640 \]

This matches our right-hand side, so:

\[ n = 19 \]

### Final Answer

Thus, the value of \( n \) is:

\[ \boxed{19} \]