How many possible values of n will make `""^(13) C_(n) lt ""^(13)C_(n+2)?`

Options

A. A) 4
B. B) 3
C. C) 6
D. D) 2

Answer

To solve the problem of how many possible values of \( n \) will make \( \binom{13}{n} < \binom{13}{n+2} \), we can follow these steps:

### Step 1: Understand the Binomial Coefficient
The binomial coefficient \( \binom{n}{r} \) represents the number of ways to choose \( r \) elements from a set of \( n \) elements. The relationship \( \binom{n}{r} = \binom{n}{n-r} \) is important because it shows that the coefficients are symmetric.

### Step 2: Set Up the Inequality
We need to find values of \( n \) such that:
\[ \binom{13}{n} < \binom{13}{n+2} \]

### Step 3: Use the Property of Binomial Coefficients
Using the property of binomial coefficients, we can express the inequality as:
\[ \frac{\binom{13}{n}}{\binom{13}{n+2}} < 1 \]
This can be rewritten using the formula for binomial coefficients:
\[ \frac{13!}{n!(13-n)!} \cdot \frac{(n+2)!(13-(n+2))!}{13!} < 1 \]
This simplifies to:
\[ \frac{(n+2)(n+1)}{(13-n)(12-n)} < 1 \]

### Step 4: Rearranging the Inequality
Rearranging gives:
\[ (n+2)(n+1) < (13-n)(12-n) \]
Expanding both sides:
\[ n^2 + 3n + 2 < 156 - 25n + n^2 \]
Cancelling \( n^2 \) from both sides:
\[ 3n + 2 < 156 - 25n \]
Combining like terms:
\[ 28n < 154 \]
Dividing by 28:
\[ n < \frac{154}{28} \approx 5.5 \]

### Step 5: Determine Possible Values of \( n \)
Since \( n \) must be a non-negative integer, the possible values for \( n \) are:
\[ n = 0, 1, 2, 3, 4, 5 \]
This gives us a total of 6 possible values.

### Conclusion
Thus, the number of possible values of \( n \) that satisfy the inequality \( \binom{13}{n} < \binom{13}{n+2} \) is:
\[ \boxed{6} \]