If `""^(n)C_(12)=""^(n)C_(6)`, then `""^(n )C_(2)` is equal to

Options

A. 72
B. 153
C. 306
D. 2556

Answer

To solve the problem, we start with the equation given in the question:

\[ {n \choose 12} = {n \choose 6} \]

### Step 1: Use the property of combinations
We know that the combination formula has the property:

\[ {n \choose r} = {n \choose n-r} \]

Applying this property, we can rewrite \( {n \choose 6} \) as:

\[ {n \choose 6} = {n \choose n-6} \]

### Step 2: Set up the equation
From the equality \( {n \choose 12} = {n \choose 6} \), we can also express \( {n \choose 12} \) using the property:

\[ {n \choose 12} = {n \choose n-12} \]

Thus, we have:

\[ {n \choose n-12} = {n \choose n-6} \]

### Step 3: Equate the two expressions
Since both expressions equal \( {n \choose 6} \), we can equate the two:

\[ n - 12 = n - 6 \]

### Step 4: Solve for \( n \)
From the equation above, we simplify:

\[ -12 = -6 \]

This is not possible, so we need to find another way to relate \( n \). Instead, we can use the fact that:

\[ 12 + 6 = n \]

Thus:

\[ n = 18 \]

### Step 5: Find \( {n \choose 2} \)
Now that we have \( n = 18 \), we can find \( {n \choose 2} \):

\[ {18 \choose 2} = \frac{18!}{2!(18-2)!} = \frac{18!}{2! \cdot 16!} \]

### Step 6: Simplify the expression
This simplifies to:

\[ {18 \choose 2} = \frac{18 \times 17 \times 16!}{2 \times 16!} = \frac{18 \times 17}{2} = \frac{306}{2} = 153 \]

### Final Answer
Thus, \( {n \choose 2} \) is equal to:

\[ \boxed{153} \]

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