9514 1404 393
Answer:
yes
Step-by-step explanation:
There are 20 ways to choose the first student, 19 ways to choose the second student, and 18 ways to choose the third student. Chosen this way, the same three students can be chosen 3×2×1 = 6 different ways. So, the number of different choices of 3 students will be ...
(20×19×18)/(3×2×1) = 1,140
There are easily enough different combinations of 3 students to have a different combination each day.
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Additional comment
The number found above is the number of combinations of 20 students taken 3 at a time: 20C3 or C(20, 3). The value can be computed from ...
nCk = n!/(k!(n-k)!)