If `""^((n-1)) C_3+""^((n-1)) C_4 gt "^n C_3` then the least value of n is

Options

A. 3
B. 4
C. 7
D. 8

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 left-hand side of the inequality:
\[ \binom{n-1}{3} + \binom{n-1}{4} \]
According to Pascal's identity, we know that:
\[ \binom{n-1}{r} + \binom{n-1}{r-1} = \binom{n}{r} \]
In our case, we can set \( r = 4 \):
\[ \binom{n-1}{4} + \binom{n-1}{3} = \binom{n}{4} \]
Thus, we can rewrite the left-hand side as:
\[ \binom{n}{4} \]

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

### Step 3: Expand the Binomial Coefficients
Next, we can express the binomial coefficients in terms of factorials:
\[ \frac{n!}{4!(n-4)!} > \frac{n!}{3!(n-3)!} \]

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

### Step 5: Simplify the Inequality
This simplifies to:
\[ \frac{1}{24(n-4)!} > \frac{1}{6(n-3)(n-4)!} \]
Cancelling \( (n-4)! \) from both sides gives:
\[ \frac{1}{24} > \frac{1}{6(n-3)} \]

### Step 6: Cross-Multiply
Cross-multiplying leads us to:
\[ 6(n-3) > 24 \]
This simplifies to:
\[ n - 3 > 4 \]
Thus:
\[ n > 7 \]

### Step 7: Find the Least Integer Value of n
The least integer value of \( n \) that satisfies this inequality is:
\[ n = 8 \]

### Final Answer
Therefore, the least value of \( n \) is \( 8 \).

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