` ""^(n-1)C_3+""^(n-1)C_4 gt ""^(n)C_3 ` if

Options

A. ` n gt 7 `
B. ` n ge 7`
C. ` n gt 6`
D. ` n ge 6`

Answer

To solve the inequality \( \binom{n-1}{3} + \binom{n-1}{4} > \binom{n}{3} \), we will follow these steps:

### Step 1: Use the Pascal's Identity
We start with the identity that states:
\[ \binom{n-1}{r} + \binom{n-1}{r+1} = \binom{n}{r+1} \]
In our case, we can set \( r = 3 \):
\[ \binom{n-1}{3} + \binom{n-1}{4} = \binom{n}{4} \]

### Step 2: Rewrite the Inequality
Now, substituting this into our inequality, we have:
\[ \binom{n}{4} > \binom{n}{3} \]

### Step 3: Express the Binomial Coefficients
Using the definition of binomial coefficients, we can express them as:
\[ \binom{n}{4} = \frac{n!}{4!(n-4)!} \]
\[ \binom{n}{3} = \frac{n!}{3!(n-3)!} \]

### Step 4: Set Up the Inequality
Substituting these into the inequality gives us:
\[ \frac{n!}{4!(n-4)!} > \frac{n!}{3!(n-3)!} \]

### Step 5: Cancel \( n! \)
Since \( n! \) is common on both sides, we can cancel it (assuming \( n \geq 4 \)):
\[ \frac{1}{4!(n-4)!} > \frac{1}{3!(n-3)!} \]

### Step 6: Cross-Multiply
Cross-multiplying gives us:
\[ 3!(n-3)! > 4!(n-4)! \]

### Step 7: Simplify the Factorials
We know that \( 4! = 4 \times 3! \), so we can simplify:
\[ 3!(n-3)! > 4 \times 3!(n-4)! \]
Dividing both sides by \( 3! \) (assuming \( n \geq 4 \)):
\[ (n-3)! > 4(n-4)! \]

### Step 8: Further Simplification
Now, we can express \( (n-3)! \) as \( (n-3)(n-4)! \):
\[ (n-3)(n-4)! > 4(n-4)! \]
Dividing both sides by \( (n-4)! \) (assuming \( n > 4 \)):
\[ n-3 > 4 \]

### Step 9: Solve for \( n \)
Finally, solving the inequality gives us:
\[ n > 7 \]

### Conclusion
Thus, the condition for the original inequality \( \binom{n-1}{3} + \binom{n-1}{4} > \binom{n}{3} \) is:
\[ n > 7 \]