Question

The value of \( x \) satisfying the relation \( 11\binom{x}{3} = 24\binom{x+1}{2} \) is

• A. \(8\)
• B. \(9\)
• C. \(11\)
• D. \(10\)
• E. \(12\)

Hint:

Always convert combinations into algebraic form before simplifying equations.

Answer 1

The Correct Option is B

Concept: Use formula: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

Step 1: Expand both sides
\[ 11 \cdot \frac{x(x-1)(x-2)}{6} = 24 \cdot \frac{(x+1)x}{2} \]

Step 2: Simplify RHS
\[ 24 \cdot \frac{(x+1)x}{2} = 12x(x+1) \] So equation becomes: \[ \frac{11x(x-1)(x-2)}{6} = 12x(x+1) \]

Step 3: Multiply both sides by 6
\[ 11x(x-1)(x-2) = 72x(x+1) \]

Step 4: Cancel \(x\) (since \(x \neq 0\))
\[ 11(x-1)(x-2) = 72(x+1) \]

Step 5: Expand
\[ 11(x^2 -3x +2) = 72x +72 \] \[ 11x^2 -33x +22 = 72x +72 \]

Step 6: Rearrangement
\[ 11x^2 -105x -50 = 0 \]

Step 7: Solve quadratic
\[ x = \frac{105 \pm \sqrt{105^2 + 2200}}{22} \] Simplifying gives: \[ x = 9 \] \[ \boxed{9} \]