Question

If \( ^nC_2 + ^nC_3 = ^6C_3 \) and \( ^nC_x = ^nC_3, x \neq 3 \), then the value of \( x \) is equal to:

• A. \( 5 \)
• B. \( 4 \)
• C. \( 2 \)
• D. \( 6 \)
• E. \( 1 \)

Hint:

Pascal's Triangle is your best friend for small values. If you forget the formula, just remember that any entry in the triangle is the sum of the two entries directly above it.

Answer 1

The Correct Option is C

Concept: We use Pascal's Identity: \( ^nC_r + ^nC_{r+1} = ^{n+1}C_{r+1} \). Additionally, we use the property of combinations: \( ^nC_x = ^nC_y \) implies either \( x = y \) or \( x + y = n \).

Step 1: Solve the first equation to find \( n \).
Using Pascal's Identity on the left side: \[ ^nC_2 + ^nC_3 = ^{n+1}C_3 \] Given: \[ ^{n+1}C_3 = ^6C_3 \] Comparing the two sides, we get \( n+1 = 6 \), which means: \[ n = 5 \]

Step 2: Apply the property for the second equation.
We are given \( ^nC_x = ^nC_3 \). Substituting \( n = 5 \): \[ ^5C_x = ^5C_3 \] From the symmetry property of combinations (\( ^nC_r = ^nC_{n-r} \)): \[ x = 3 \quad \text{or} \quad x = 5 - 3 = 2 \]

Step 3: Select the correct value for \( x \).
The problem explicitly states that \( x \neq 3 \). Therefore, \( x = 2 \).