If `(n-1) C_6 + (n-1)C_7 > nC_6` then

Options

A. `ngt4`
B. `ngt12`
C. `nge13`
D. `ngt13`

Answer

To solve the inequality \( (n-1)C_6 + (n-1)C_7 > nC_6 \), we will use the properties of combinations. Let's break it down step by step.

### Step 1: Rewrite the Combinations
We start with the given inequality:
\[ (n-1)C_6 + (n-1)C_7 > nC_6 \]

### Step 2: Apply the Combination Property
Using the property of combinations, we know that:
\[ (n-1)C_r + (n-1)C_{r-1} = nC_r \]
For \( r = 7 \), we can rewrite \( (n-1)C_6 + (n-1)C_7 \) as:
\[ nC_7 \]
Thus, we can rewrite our inequality as:
\[ nC_7 > nC_6 \]

### Step 3: Write the Combinations in Factorial Form
Now, we express \( nC_7 \) and \( nC_6 \) in terms of factorials:
\[ nC_7 = \frac{n!}{7!(n-7)!} \]
\[ nC_6 = \frac{n!}{6!(n-6)!} \]

### Step 4: Set Up the Inequality
Now substituting these into our inequality gives:
\[ \frac{n!}{7!(n-7)!} > \frac{n!}{6!(n-6)!} \]

### Step 5: Cancel \( n! \)
Since \( n! \) is common on both sides, we can cancel it (assuming \( n! \neq 0 \)):
\[ \frac{1}{7!(n-7)!} > \frac{1}{6!(n-6)!} \]

### Step 6: Cross Multiply
Cross multiplying gives:
\[ 6!(n-6)! > 7!(n-7)! \]

### Step 7: Simplify the Inequality
Now we can simplify:
\[ 6!(n-6)! > 7 \cdot 6!(n-7)! \]
Dividing both sides by \( 6! \) (assuming \( 6! \neq 0 \)):
\[ (n-6)! > 7(n-7)! \]

### Step 8: Further Simplify
We can express \( (n-6)! \) as:
\[ (n-6)(n-7)! > 7(n-7)! \]
Dividing both sides by \( (n-7)! \) (assuming \( (n-7)! \neq 0 \)):
\[ n-6 > 7 \]

### Step 9: Solve for \( n \)
Adding 6 to both sides gives:
\[ n > 13 \]

### Conclusion
Thus, the solution to the inequality is:
\[ n > 13 \]