The least positive integer n such that `((n-1),(3)) + ((n-1),(4)) gt ((n),(3))` is

Options

A. 6
B. 7
C. 8
D. 9

Answer

To solve the problem, we need to find the least positive integer \( n \) such that:

\[ \binom{n-1}{3} + \binom{n-1}{4} > \binom{n}{3} \]

### Step 1: Use the Pascal's Identity
We can use the identity:

\[ \binom{n-1}{r} + \binom{n-1}{r+1} = \binom{n}{r+1} \]

In our case, we can set \( r = 3 \):

\[ \binom{n-1}{3} + \binom{n-1}{4} = \binom{n}{4} \]

### Step 2: Rewrite the Inequality
Substituting this into our inequality gives us:

\[ \binom{n}{4} > \binom{n}{3} \]

### Step 3: Express the Binomial Coefficients
Using the formula for binomial coefficients, we can express these:

\[ \binom{n}{4} = \frac{n!}{4!(n-4)!} \quad \text{and} \quad \binom{n}{3} = \frac{n!}{3!(n-3)!} \]

### Step 4: Set Up the Inequality
Now we set up the inequality:

\[ \frac{n!}{4!(n-4)!} > \frac{n!}{3!(n-3)!} \]

### Step 5: Cancel \( n! \)
Since \( n! \) is common on both sides and \( n > 4 \), we can cancel it:

\[ \frac{1}{4!(n-4)!} > \frac{1}{3!(n-3)!} \]

### Step 6: Simplify the Inequality
Rearranging gives us:

\[ 3!(n-3)! > 4!(n-4)! \]

### Step 7: Substitute Factorials
Substituting \( 3! = 6 \) and \( 4! = 24 \):

\[ 6(n-3)! > 24(n-4)! \]

### Step 8: Cancel \( (n-4)! \)
We can cancel \( (n-4)! \) since \( n > 4 \):

\[ 6(n-3) > 24 \]

### Step 9: Solve for \( n \)
Dividing both sides by 6:

\[ n-3 > 4 \]

Adding 3 to both sides gives:

\[ n > 7 \]

### Step 10: Find the Least Positive Integer
The least positive integer \( n \) that satisfies this inequality is:

\[ n = 8 \]

Thus, the answer is:

\[ \boxed{8} \]

---