If `""^(n)C_(2)= ""^(n)C_(3)`, then n = ……….

Answer

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

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

### Step 2: Set the two combinations equal to each other
Now, we set the two expressions equal:
\[ \frac{n!}{2!(n-2)!} = \frac{n!}{3!(n-3)!} \]

### Step 3: Cancel \( n! \) from both sides
Since \( n! \) is common in both expressions, we can cancel it (assuming \( n \geq 3 \)):
\[ \frac{1}{2!(n-2)!} = \frac{1}{3!(n-3)!} \]

### Step 4: Cross-multiply
Cross-multiplying gives us:
\[ 3!(n-3)! = 2!(n-2)! \]

### Step 5: Substitute the factorial values
We know that \( 3! = 6 \) and \( 2! = 2 \), so we substitute these values:
\[ 6(n-3)! = 2(n-2)! \]

### Step 6: Express \( (n-2)! \) in terms of \( (n-3)! \)
We can express \( (n-2)! \) as:
\[ (n-2)! = (n-2)(n-3)! \]
Substituting this into our equation gives:
\[ 6(n-3)! = 2(n-2)(n-3)! \]

### Step 7: Cancel \( (n-3)! \) from both sides
Assuming \( n \geq 3 \), we can cancel \( (n-3)! \):
\[ 6 = 2(n-2) \]

### Step 8: Solve for \( n \)
Now, divide both sides by 2:
\[ 3 = n - 2 \]
Adding 2 to both sides gives:
\[ n = 5 \]

Thus, the value of \( n \) is \( 5 \).

### Final Answer:
n = 5