anonymous
  • anonymous
Counting principles
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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kropot72
  • 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 \[\large \frac{9!}{3!3!3!}\]
kropot72
  • kropot72
Yes, I believe it is basically the same.
anonymous
  • anonymous
Thank you so much @kropot72 =)

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kropot72
  • kropot72
You're welcome :)

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