Question

If \( \frac{C(2n,3)}{C(n,2)} = \frac{44}{3} \), then \( n \) is equal to

• A. 6
• B. 7
• C. 5
• D. 8

Hint:

Cancel common factors like \((n-1)\) early to simplify quickly.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The question asks to find the value of \( n \) given a ratio of combinations. We expand the combination formula \( {}^nC_r = \frac{n!}{r!(n-r)!} \).
: Key Formula or Approach:
\[ \frac{{}^{2n}C_3}{{}^{n}C_2} = \frac{44}{3} \] Step 2: Detailed Explanation:
Expand the combinations: \[ \frac{\frac{(2n)(2n-1)(2n-2)}{3 \cdot 2 \cdot 1}}{\frac{n(n-1)}{2 \cdot 1}} = \frac{44}{3} \] Simplify the fractions: \[ \frac{\frac{(2n)(2n-1) \cdot 2(n-1)}{6}}{\frac{n(n-1)}{2}} = \frac{44}{3} \] \[ \frac{2 \cdot 2n(2n-1)(n-1)}{6 \cdot n(n-1)} = \frac{44}{3} \] The terms \( n \) and \( (n-1) \) cancel out: \[ \frac{4(2n-1)}{6} = \frac{44}{3} \] \[ \frac{2(2n-1)}{3} = \frac{44}{3} \] Cross-multiply or cancel 3: \[ 2(2n-1) = 44 \] \[ 2n - 1 = 22 \] \[ 2n = 23 \] Wait, let's re-check calculation. From the ratio \( \frac{2n(2n-1)2(n-1)}{6} \times \frac{2}{n(n-1)} = \frac{44}{3} \).
\[ \frac{4n(2n-1)(n-1)}{3n(n-1)} = \frac{44}{3} \] \[ \frac{4(2n-1)}{3} = \frac{44}{3} \] \[ 4(2n-1) = 44 \implies 2n - 1 = 11 \implies 2n = 12 \implies n = 6 \].
Step 3: Final Answer:
The value of n is 6.