Question

If \( ^nC_3 = 220 \), then \( n = \ ? \)

• A. 11
• B. 12
• C. 10
• D. 9

Hint:

When solving \( ^nC_r = \text{value} \), always use the formula \( \frac{n(n-1)(n-2)...}{r!} \) and try small values of \( n \) by trial for small \( r \), especially when the RHS is a manageable number.

Answer 1

The Correct Option is B

We are given: \[ ^nC_3 = 220 \] Step 1: Use the formula for combinations
The general formula for combinations is: \[ ^nC_r = \frac{n!}{r!(n-r)!} \] For \( r = 3 \), the formula becomes: \[ ^nC_3 = \frac{n(n-1)(n-2)}{6} \] Step 2: Substitute the value given in the question
\[ \frac{n(n-1)(n-2)}{6} = 220 \] Step 3: Multiply both sides by 6 to eliminate the denominator
\[ n(n-1)(n-2) = 1320 \] Step 4: Solve the cubic equation by trial
Try \( n = 12 \): \[ 12 \times 11 \times 10 = 1320 \quad \text{✓} \] So, the required value of \( n \) is: \[ \boxed{12} \]