If `""^(n)C_(3) + ""^(n)C_4 gt ""^(n+1) C_3 , ` then.

Options

A. ` n gt 6`
B. ` n gt 7 `
C. ` n lt 6`
D. none of these

Answer

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

### Step 1: Rewrite the left-hand side
We start with the left-hand side of the inequality:
\[ \binom{n}{3} + \binom{n}{4} \]
Using the identity \( \binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1} \), we can rewrite this as:
\[ \binom{n+1}{4} \]

### Step 2: Rewrite the right-hand side
Now, we look at the right-hand side:
\[ \binom{n+1}{3} \]

### Step 3: Set up the inequality
We now have:
\[ \binom{n+1}{4} > \binom{n+1}{3} \]

### Step 4: Express both sides using factorials
Using the definition of combinations, we can express both sides:
\[ \frac{(n+1)!}{4!(n+1-4)!} > \frac{(n+1)!}{3!(n+1-3)!} \]
This simplifies to:
\[ \frac{(n+1)!}{4!(n-3)!} > \frac{(n+1)!}{3!(n-2)!} \]

### Step 5: Cancel the common terms
Since \( (n+1)! \) appears in both sides, we can cancel it out (assuming \( n+1 \neq 0 \)):
\[ \frac{1}{4!(n-3)!} > \frac{1}{3!(n-2)!} \]

### Step 6: Cross-multiply
Cross-multiplying gives us:
\[ 3!(n-2)! > 4!(n-3)! \]
Substituting \( 3! = 6 \) and \( 4! = 24 \):
\[ 6(n-2)! > 24(n-3)! \]

### Step 7: Simplify the inequality
We can simplify \( (n-2)! \) and \( (n-3)! \):
\[ 6(n-2)(n-3)! > 24(n-3)! \]
Dividing both sides by \( (n-3)! \) (assuming \( n-3 \neq 0 \)):
\[ 6(n-2) > 24 \]

### Step 8: Solve for \( n \)
Dividing both sides by 6:
\[ n-2 > 4 \]
Thus:
\[ n > 6 \]

### Conclusion
The solution to the inequality is \( n > 6 \).