Question

How many combinations of 3 students can a teacher choose from 32 students?   A. 96   B. 4960   C. 29,760   D. 32,768

Answer 1

${32\choose3}=\frac{32!}{3!*(32-3)!}=\frac{30*31*32}{6}=4960$

Answer 2

$\frac{32*31*30}{3*2*1}$

The answer is:$\frac{29760}{6}$ or $4960$