Answer:
B
Step-by-step explanation:
Using the definition
n ! = n(n - 1)(n - 2).... × 3 × 2 × 1
Then
[tex]\frac{9!}{6!3!}[/tex]
= [tex]\frac{9(8)(7)(6)(5)(4)(3)(2)(1)}{6(5)(4)(3)(2)(1)3(2)(1)}[/tex]
Cancel 6(5)(4)(3)(2)(1) on numerator/ denominator, leaving
[tex]\frac{9(8)(7)}{3(2)(1)}[/tex]
= [tex]\frac{504}{6}[/tex]
= 84 → B
