If `""^(n-1)C_(3)+""^(n-1)C_(4)gt""^nC_(3)`, then

Options

A. `nle4`
B. `ngt5`
C. `ngt7`
D. none of these

Answer

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

### Step 1: Rewrite the Left Side
Using the identity for combinations, we know that:
\[ \binom{n-1}{r} + \binom{n-1}{r+1} = \binom{n}{r+1} \]
For our case, we can set \( r = 3 \):
\[ \binom{n-1}{3} + \binom{n-1}{4} = \binom{n}{4} \]

### Step 2: Substitute into the Inequality
Now, we can substitute this back into our original inequality:
\[ \binom{n}{4} > \binom{n}{3} \]

### Step 3: Write the Combinations in Factorial Form
Using the formula for combinations, we can express both sides:
\[ \frac{n!}{4!(n-4)!} > \frac{n!}{3!(n-3)!} \]

### Step 4: Simplify the Inequality
Since \( n! \) is common on both sides, we can cancel it (assuming \( n > 0 \)):
\[ \frac{1}{4!(n-4)!} > \frac{1}{3!(n-3)!} \]
This simplifies to:
\[ 3!(n-3)! > 4!(n-4)! \]

### Step 5: Further Simplify
We can express \( 4! \) as \( 4 \times 3! \):
\[ 3!(n-3)! > 4 \times 3!(n-4)! \]
Now, we can cancel \( 3! \) from both sides:
\[ (n-3)! > 4(n-4)! \]

### Step 6: Rewrite \( (n-3)! \)
We can express \( (n-3)! \) as \( (n-3)(n-4)! \):
\[ (n-3)(n-4)! > 4(n-4)! \]
Assuming \( (n-4)! \neq 0 \) (which is true for \( n > 4 \)), we can cancel \( (n-4)! \):
\[ n-3 > 4 \]

### Step 7: Solve for \( n \)
This simplifies to:
\[ n > 7 \]

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