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:
\[ \binom{n+1}{6} - \binom{n}{4} > \binom{n}{5} \]

### Step 2: Use Pascal's Identity
Using Pascal's identity, we know that:
\[ \binom{n+1}{6} = \binom{n}{6} + \binom{n}{5} \]
Thus, we can rewrite the left side:
\[ \binom{n}{6} + \binom{n}{5} - \binom{n}{4} > \binom{n}{5} \]

### Step 3: Simplify the Inequality
Now, we can simplify the inequality:
\[ \binom{n}{6} + \binom{n}{5} - \binom{n}{4} - \binom{n}{5} > 0 \]
This simplifies to:
\[ \binom{n}{6} - \binom{n}{4} > 0 \]

### Step 4: Express the Binomial Coefficients
We can express the binomial coefficients:
\[ \binom{n}{6} = \frac{n!}{6!(n-6)!} \quad \text{and} \quad \binom{n}{4} = \frac{n!}{4!(n-4)!} \]

### Step 5: Set Up the Inequality
Now we need to set up the inequality:
\[ \frac{n!}{6!(n-6)!} > \frac{n!}{4!(n-4)!} \]
We can cancel \( n! \) from both sides (assuming \( n \geq 6 \)):
\[ \frac{1}{6!(n-6)!} > \frac{1}{4!(n-4)!} \]

### Step 6: Cross Multiply
Cross multiplying gives us:
\[ 4!(n-4)! > 6!(n-6)! \]
This simplifies to:
\[ 24(n-4)(n-5) > 720 \]

### Step 7: Solve the Inequality
Dividing both sides by 24:
\[ (n-4)(n-5) > 30 \]
Expanding gives:
\[ n^2 - 9n + 20 > 30 \]
Rearranging gives:
\[ n^2 - 9n - 10 > 0 \]

### Step 8: Factor the Quadratic
Factoring the quadratic:
\[ (n-10)(n+1) > 0 \]

### Step 9: Determine the Solution
The roots of the equation are \( n = 10 \) and \( n = -1 \). The inequality \( (n-10)(n+1) > 0 \) holds true for:
- \( n < -1 \) or \( n > 10 \)

Since we are looking for \( n > 10 \), the solution is:
\[ n > 10 \]

### Final Answer
Thus, the inequality holds true for all \( n \) greater than **10**.

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