IF `""^n C_12= "^n C_8` then n=

Options

A. 20
B. 12
C. 6
D. 30

Answer

To solve the equation \( \binom{n}{12} = \binom{n}{8} \), we can follow these steps:

### Step 1: Write the formula for combinations
The formula for combinations is given by:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Using this formula, we can express both sides of the equation.

### Step 2: Set up the equation
From the equation \( \binom{n}{12} = \binom{n}{8} \), we can write:
\[ \frac{n!}{12!(n-12)!} = \frac{n!}{8!(n-8)!} \]

### Step 3: Cancel \( n! \) from both sides
Since \( n! \) is common on both sides, we can cancel it out:
\[ \frac{1}{12!(n-12)!} = \frac{1}{8!(n-8)!} \]

### Step 4: Cross-multiply
Cross-multiplying gives us:
\[ 8!(n-12)! = 12!(n-8)! \]

### Step 5: Rewrite \( 12! \)
We can rewrite \( 12! \) as \( 12 \times 11 \times 10 \times 9 \times 8! \):
\[ 8!(n-12)! = 12 \times 11 \times 10 \times 9 \times 8!(n-8)! \]

### Step 6: Cancel \( 8! \)
Now, we can cancel \( 8! \) from both sides:
\[ (n-12)! = 12 \times 11 \times 10 \times 9 \times (n-8)! \]

### Step 7: Rewrite \( (n-12)! \)
We can express \( (n-12)! \) in terms of \( (n-8)! \):
\[ (n-12)! = \frac{(n-8)!}{(n-8)(n-9)(n-10)(n-11)} \]

### Step 8: Set the equation
Substituting back, we have:
\[ \frac{(n-8)!}{(n-8)(n-9)(n-10)(n-11)} = 12 \times 11 \times 10 \times 9 \times (n-8)! \]

### Step 9: Cancel \( (n-8)! \)
Cancelling \( (n-8)! \) from both sides gives:
\[ \frac{1}{(n-8)(n-9)(n-10)(n-11)} = 12 \times 11 \times 10 \times 9 \]

### Step 10: Cross-multiply
Cross-multiplying leads to:
\[ 1 = 12 \times 11 \times 10 \times 9 \times (n-8)(n-9)(n-10)(n-11) \]

### Step 11: Solve for \( n \)
Now, we can simplify and solve for \( n \). From the earlier steps, we can also deduce that:
\[ 12 = n - 8 \]
Thus,
\[ n = 20 \]

### Final Answer
The value of \( n \) is \( 20 \).

---