Hi,
When chosen, the order of each group of 3 students does not matter, so we use combinations.
[tex]\Big{\binom{16}{3}}=\dfrac{16!}{3!\cdot(16-13)!}=\dfrac{16\cdot 15\cdot 14\cdot 13!}{1\cdot 2\cdot 3\cdot 13!}=\dfrac{16\cdot 15\cdot 14}{6}=8\cdot 5\cdot 14=560\;combinations.[/tex]
Green eyes.
