IF `""^(n+1)C_3=2."^n C_2` then n=

Options

A. 3
B. 4
C. 5
D. 6

Answer

To solve the equation \( \binom{n+1}{3} = 2 \cdot \binom{n}{2} \), we will follow these steps:

### Step 1: Write the combinations in terms of factorials.
The combination formula is given by:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Using this, we can express \( \binom{n+1}{3} \) and \( \binom{n}{2} \):
\[ \binom{n+1}{3} = \frac{(n+1)!}{3!(n+1-3)!} = \frac{(n+1)!}{3!(n-2)!} \]
\[ \binom{n}{2} = \frac{n!}{2!(n-2)!} \]

### Step 2: Substitute the combinations into the equation.
Now substituting the combinations into the equation:
\[ \frac{(n+1)!}{3!(n-2)!} = 2 \cdot \frac{n!}{2!(n-2)!} \]

### Step 3: Simplify the equation.
We can cancel \( (n-2)! \) from both sides:
\[ \frac{(n+1)!}{3!} = 2 \cdot \frac{n!}{2!} \]
Now, recall that \( (n+1)! = (n+1) \cdot n! \):
\[ \frac{(n+1) \cdot n!}{3!} = 2 \cdot \frac{n!}{2!} \]
We can cancel \( n! \) from both sides (assuming \( n! \neq 0 \)):
\[ \frac{(n+1)}{3!} = 2 \cdot \frac{1}{2!} \]

### Step 4: Calculate the factorials.
Calculating \( 3! \) and \( 2! \):
\[ 3! = 6 \quad \text{and} \quad 2! = 2 \]
So, we have:
\[ \frac{(n+1)}{6} = 2 \cdot \frac{1}{2} = 1 \]

### Step 5: Solve for \( n \).
Now, multiply both sides by 6:
\[ n + 1 = 6 \]
Subtracting 1 from both sides gives:
\[ n = 5 \]

### Final Answer:
Thus, the value of \( n \) is \( 5 \).

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