## anonymous one year ago Counting principles

1. kropot72

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 $\large \frac{9!}{3!3!3!}$

2. kropot72

Yes, I believe it is basically the same.

3. anonymous

Thank you so much @kropot72 =)

4. kropot72

You're welcome :)