Answer:
15
Step-by-step explanation:
Using the definition of n[tex]C_{r}[/tex]
n[tex]C_{r}[/tex] = [tex]\frac{n!}{r!(n-r)!}[/tex]
where n! = n(n - 1)(n - 2) × 3 × 2 × 1
Hence
6[tex]C_{2}[/tex]
= [tex]\frac{6!}{2!(4!)}[/tex]
= [tex]\frac{6(5)(4)(3)(2)(1)}{2(1)4(3)(2)(1)}[/tex]
Cancel 4(3)(2)(1) on numerator/ denominator, leaving
= [tex]\frac{6(5)}{2(1)}[/tex] = [tex]\frac{30}{2}[/tex] = 15
