If `C(n, 7)=C(n, 5)`, Find the value of n?

Options

A. 15
B. 12
C. 18
D. 2

Answer

To solve the equation \( C(n, 7) = C(n, 5) \), we can follow these steps:

### Step 1: Write the combinations in terms of factorials
The combination formula is given by:
\[ C(n, r) = \frac{n!}{r!(n-r)!} \]
So, we can write:
\[ C(n, 7) = \frac{n!}{7!(n-7)!} \]
\[ C(n, 5) = \frac{n!}{5!(n-5)!} \]

### Step 2: Set the two combinations equal to each other
From the problem statement, we have:
\[ \frac{n!}{7!(n-7)!} = \frac{n!}{5!(n-5)!} \]

### Step 3: Cancel \( n! \) from both sides
Since \( n! \) is common in both terms, we can cancel it:
\[ \frac{1}{7!(n-7)!} = \frac{1}{5!(n-5)!} \]

### Step 4: Cross-multiply to eliminate the fractions
Cross-multiplying gives us:
\[ 5!(n-5)! = 7!(n-7)! \]

### Step 5: Substitute \( 7! \) and \( 5! \)
Recall that \( 7! = 7 \times 6 \times 5! \), so we can rewrite the equation:
\[ 5!(n-5)! = 7 \times 6 \times 5!(n-7)! \]
Now, we can cancel \( 5! \) from both sides:
\[ (n-5)! = 7 \times 6 \times (n-7)! \]

### Step 6: Rewrite \( (n-5)! \) in terms of \( (n-7)! \)
We can express \( (n-5)! \) as:
\[ (n-5)! = (n-5)(n-6)(n-7)! \]
Substituting this into the equation gives:
\[ (n-5)(n-6)(n-7)! = 42(n-7)! \]

### Step 7: Cancel \( (n-7)! \) from both sides
Assuming \( n-7 \neq 0 \) (which means \( n \neq 7 \)), we can cancel \( (n-7)! \):
\[ (n-5)(n-6) = 42 \]

### Step 8: Expand and rearrange the equation
Expanding the left side gives:
\[ n^2 - 11n + 30 = 42 \]
Rearranging gives:
\[ n^2 - 11n - 12 = 0 \]

### Step 9: Factor the quadratic equation
Factoring the quadratic:
\[ (n - 12)(n + 1) = 0 \]

### Step 10: Solve for \( n \)
Setting each factor to zero gives:
\[ n - 12 = 0 \quad \Rightarrow \quad n = 12 \]
\[ n + 1 = 0 \quad \Rightarrow \quad n = -1 \quad (\text{not valid since } n \text{ must be positive}) \]

Thus, the solution is:
\[ n = 12 \]

### Final Answer:
The value of \( n \) is \( 12 \).

---