Question

Counting principles

Answer 1

Let the three tests be A, B and C.
There are 9 choices for the first student to take test A, 8 choices for the second student to take test A, and 7 choices for the third student to take test A. This would give a total of
\[\large \frac{9\times8\times7}{3!}\]
ways for test A. Note the reason for dividing by factorial 3 is that the order of selection does not matter.
Using the same reasoning as for test A, the number of ways for test B is given by
\[\large \frac{6\times5\times4}{3!}\]
and for test C
\[\large \frac{3\times2\times1}{3!}\]
Therefore the total number of ways of selecting the students is given by
\[\large \frac{9!}{3!3!3!}\]

Answer 2

Yes, I believe it is basically the same.

Answer 3

Thank you so much @kropot72 =)

Answer 4

You're welcome :)