If `.^(20)C_(n+1)=.^(n)C_(16)`, the value of n is

Options

A. 7
B. 10
C. 13
D. None of these

Answer

To solve the equation \( \binom{20}{n+1} = \binom{n}{16} \), we will follow these steps:

### Step 1: Understand the Binomial Coefficient
The binomial coefficient \( \binom{n}{r} \) is defined as:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
In our case, we have \( \binom{20}{n+1} \) and \( \binom{n}{16} \).

### Step 2: Set Up the Equation
From the equation \( \binom{20}{n+1} = \binom{n}{16} \), we can express it using the formula for binomial coefficients:
\[ \frac{20!}{(n+1)!(20-n-1)!} = \frac{n!}{16!(n-16)!} \]

### Step 3: Analyze the Conditions
We know that for \( \binom{20}{n+1} \) to be valid, \( n+1 \) must be less than or equal to 20, which gives us:
\[ n + 1 \leq 20 \implies n \leq 19 \]
For \( \binom{n}{16} \) to be valid, \( n \) must be at least 16:
\[ n \geq 16 \]

### Step 4: Combine the Conditions
From the above conditions, we have:
\[ 16 \leq n \leq 19 \]
This means \( n \) can take the values 16, 17, 18, or 19.

### Step 5: Check Possible Values
Now we will check which of these values satisfy the original equation \( \binom{20}{n+1} = \binom{n}{16} \):

1. **For \( n = 16 \)**:
\[ \binom{20}{17} = \binom{16}{16} \implies 20 = 1 \quad \text{(not valid)} \]

2. **For \( n = 17 \)**:
\[ \binom{20}{18} = \binom{17}{16} \implies 190 = 17 \quad \text{(not valid)} \]

3. **For \( n = 18 \)**:
\[ \binom{20}{19} = \binom{18}{16} \implies 20 = 153 \quad \text{(not valid)} \]

4. **For \( n = 19 \)**:
\[ \binom{20}{20} = \binom{19}{16} \implies 1 = 969 \quad \text{(not valid)} \]

### Conclusion
None of the values from 16 to 19 satisfy the equation. Therefore, the answer is that there is no valid \( n \) in the given range.

### Final Answer
The value of \( n \) is **none of these**.

---