If 3(⁶C₃) = 10(^xC₂), then x = ?

Concept:

Combinatorics:

• ⁿC_r = \(\rm \frac {n!}{r!(n\ -\ r)!}\).
• n! = 1 × 2 × 3 × ... × n.
• 0! = 1.

Calculation:

Given equation is: 3(6C3) = 10(xC2)

⇒ \(\rm 3×\left[\frac {6!}{3!(6\ -\ 3)!}\right]=10×\left[\frac {x!}{2!(x\ -\ 2)!}\right]\)

⇒ \(\rm 3×\left(\frac {6\ ×\ 5\ ×\ 4}{3\ ×\ 2\ ×\ 1}\right)=10×\left[\frac {x(x\ -\ 1)}{2\ ×\ 1}\right]\)

⇒ x(x - 1) = 12 = 4 × 3

⇒ x = 4.