A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
Counting principles
anonymous
 one year ago
Counting principles

This Question is Closed

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1Let 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 \[\large \frac{9!}{3!3!3!}\]

kropot72
 one year ago
Best ResponseYou've already chosen the best response.1Yes, I believe it is basically the same.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much @kropot72 =)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.