`""^(n+1) C_6+""^n C_4gt ""^(n+2) C_5-"^n C_5` will hold good for all n greater than

Options

A. 1
B. 10
C. 9
D. 8

Answer

To solve the inequality \( \binom{n+1}{6} + \binom{n}{4} > \binom{n+2}{5} - \binom{n}{5} \), we will follow these steps:

### Step 1: Rewrite the Inequality
Start by rewriting the inequality in a more manageable form:
\[ \binom{n+1}{6} + \binom{n}{4} > \binom{n+2}{5} - \binom{n}{5} \]

### Step 2: Use Pascal's Identity
Recall Pascal's identity:
\[ \binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1} \]
We can apply this identity to the left-hand side:
\[ \binom{n+1}{6} + \binom{n}{4} = \binom{n+1}{6} + \binom{n}{4} = \binom{n+1}{5} + \binom{n}{4} \]
This gives us:
\[ \binom{n+1}{5} + \binom{n}{4} > \binom{n+2}{5} - \binom{n}{5} \]

### Step 3: Simplify the Right-Hand Side
Now, let's simplify the right-hand side:
Using Pascal's identity again:
\[ \binom{n+2}{5} = \binom{n+1}{5} + \binom{n+1}{4} \]
Thus, we can rewrite the right-hand side as:
\[ \binom{n+1}{5} + \binom{n+1}{4} - \binom{n}{5} \]

### Step 4: Combine Both Sides
Now we have:
\[ \binom{n+1}{5} + \binom{n}{4} > \binom{n+1}{5} + \binom{n+1}{4} - \binom{n}{5} \]
Subtracting \(\binom{n+1}{5}\) from both sides gives:
\[ \binom{n}{4} > \binom{n+1}{4} - \binom{n}{5} \]

### Step 5: Use Pascal's Identity Again
Using Pascal's identity on the right-hand side:
\[ \binom{n+1}{4} = \binom{n}{4} + \binom{n}{3} \]
Thus, we can rewrite the inequality as:
\[ \binom{n}{4} > \left(\binom{n}{4} + \binom{n}{3}\right) - \binom{n}{5} \]
This simplifies to:
\[ 0 > \binom{n}{3} - \binom{n}{5} \]

### Step 6: Rearrange the Inequality
Rearranging gives:
\[ \binom{n}{5} > \binom{n}{3} \]

### Step 7: Analyze the Inequality
The inequality \(\binom{n}{5} > \binom{n}{3}\) holds true for \(n \geq 9\). This is because the binomial coefficient \(\binom{n}{k}\) increases as \(n\) increases for fixed \(k\).

### Conclusion
Thus, the inequality \( \binom{n+1}{6} + \binom{n}{4} > \binom{n+2}{5} - \binom{n}{5} \) holds for all \( n > 9 \).