Find the value of n when `""^(n)C_(2)=105` ?

Answer

To find the value of \( n \) such that \( \binom{n}{2} = 105 \), we can follow these steps:

### Step 1: Write the combination formula
The formula for combinations is given by:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
For our case, \( r = 2 \):
\[ \binom{n}{2} = \frac{n!}{2!(n-2)!} \]

### Step 2: Simplify the formula
Substituting \( r = 2 \) into the formula, we have:
\[ \binom{n}{2} = \frac{n(n-1)}{2} \]
Setting this equal to 105 gives:
\[ \frac{n(n-1)}{2} = 105 \]

### Step 3: Eliminate the fraction
To eliminate the fraction, multiply both sides by 2:
\[ n(n-1) = 210 \]

### Step 4: Rearrange the equation
Rearranging the equation gives us:
\[ n^2 - n - 210 = 0 \]

### Step 5: Factor the quadratic equation
We need to factor the quadratic equation \( n^2 - n - 210 = 0 \). We look for two numbers that multiply to \(-210\) and add to \(-1\). The numbers are \( -15 \) and \( 14 \):
\[ (n - 15)(n + 14) = 0 \]

### Step 6: Solve for \( n \)
Setting each factor to zero gives us:
1. \( n - 15 = 0 \) → \( n = 15 \)
2. \( n + 14 = 0 \) → \( n = -14 \)

Since \( n \) must be a non-negative integer, we discard \( n = -14 \).

### Final Answer
Thus, the value of \( n \) is:
\[ \boxed{15} \]

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