Counting principles

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Counting principles

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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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 \[\large \frac{9!}{3!3!3!}\]
Yes, I believe it is basically the same.
Thank you so much @kropot72 =)

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You're welcome :)

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