what is the least possible value of n if :
` ""^(n-1)C_3 +^(n-1)C_4 gt ""^nC_3`

Options

A. 6
B. 8
C. 7
D. 12

Answer

To solve the problem, we need to find the least possible value of \( n \) such that:

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

### Step 1: Use the Binomial Coefficient Identity
We can use the identity for binomial coefficients:

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

In our case, we can rewrite:

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

### Step 2: Substitute into the Inequality
Now, substituting this into our inequality gives us:

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

### Step 3: Write the Binomial Coefficients
We can express the binomial coefficients as follows:

\[ \binom{n}{4} = \frac{n(n-1)(n-2)(n-3)}{4!} = \frac{n(n-1)(n-2)(n-3)}{24} \]

\[ \binom{n}{3} = \frac{n(n-1)(n-2)}{3!} = \frac{n(n-1)(n-2)}{6} \]

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

\[ \frac{n(n-1)(n-2)}{24} > \frac{n(n-1)(n-2)}{6} \]

### Step 5: Simplify the Inequality
To simplify, we can multiply both sides by 24 (assuming \( n(n-1)(n-2) \) is positive):

\[ n(n-1)(n-2) > 4n(n-1)(n-2) \]

### Step 6: Rearranging the Terms
This simplifies to:

\[ 0 > 3n(n-1)(n-2) \]

This means \( n(n-1)(n-2) < 0 \). The product \( n(n-1)(n-2) \) is negative when \( n \) is less than 2.

### Step 7: Testing Values of n
Since we want the least possible value of \( n \), we can start testing values:

- For \( n = 6 \):
\[ \binom{5}{3} + \binom{5}{4} = 10 + 5 = 15 \]
\[ \binom{6}{3} = 20 \quad \text{(not satisfied)} \]

- For \( n = 7 \):
\[ \binom{6}{3} + \binom{6}{4} = 20 + 15 = 35 \]
\[ \binom{7}{3} = 35 \quad \text{(not satisfied)} \]

- For \( n = 8 \):
\[ \binom{7}{3} + \binom{7}{4} = 35 + 35 = 70 \]
\[ \binom{8}{3} = 56 \quad \text{(satisfied)} \]

### Conclusion
The least possible value of \( n \) that satisfies the inequality is:

\[ \boxed{8} \]