The inequality `.^(n+1)C_(6)-.^(n)C_(4) gt .^(n)C_(5)` holds true for all n greater than ________.

Options

A. 8
B. 9
C. 7
D. 10

Answer

To solve the inequality \( \binom{n+1}{6} - \binom{n}{4} > \binom{n}{5} \), we will follow these steps:

### Step 1: Rewrite the Inequality
We start with the given inequality:
\[ \binom{n+1}{6} - \binom{n}{4} > \binom{n}{5} \]
We can rearrange this to:
\[ \binom{n+1}{6} > \binom{n}{5} + \binom{n}{4} \]

### Step 2: Apply the Combination Identity
Using the combination identity \( \binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r} \), we can simplify the right-hand side:
\[ \binom{n}{5} + \binom{n}{4} = \binom{n+1}{5} \]
Thus, our inequality becomes:
\[ \binom{n+1}{6} > \binom{n+1}{5} \]

### Step 3: Apply the Combination Formula
Using the formula for combinations, we can write:
\[ \binom{n+1}{6} = \frac{(n+1)!}{6!(n-5)!} \quad \text{and} \quad \binom{n+1}{5} = \frac{(n+1)!}{5!(n-4)!} \]
Now substituting these into our inequality gives:
\[ \frac{(n+1)!}{6!(n-5)!} > \frac{(n+1)!}{5!(n-4)!} \]

### Step 4: Cancel Common Terms
Since \( (n+1)! \) is common on both sides, we can cancel it out (assuming \( n+1 \neq 0 \)):
\[ \frac{1}{6!(n-5)!} > \frac{1}{5!(n-4)!} \]

### Step 5: Simplify the Inequality
This simplifies to:
\[ 5!(n-4)! > 6!(n-5)! \]
We know \( 6! = 6 \times 5! \), so we can rewrite the inequality as:
\[ (n-4)! > 6(n-5)! \]
Now, using the factorial property \( (n-4)! = (n-4)(n-5)! \), we can substitute:
\[ (n-4)(n-5)! > 6(n-5)! \]

### Step 6: Cancel \( (n-5)! \)
Assuming \( (n-5)! \neq 0 \) (which is true for \( n > 5 \)), we can cancel \( (n-5)! \):
\[ n-4 > 6 \]

### Step 7: Solve for \( n \)
Rearranging gives:
\[ n > 10 \]

### Conclusion
The inequality \( \binom{n+1}{6} - \binom{n}{4} > \binom{n}{5} \) holds true for all \( n > 10 \).