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(a) if any student may serve in either of these roles but no student may serve in multiple roles and (b) if any student may serve in multiple roles?

here for the a, i know the answer as 657,720.

for b if I use the combination formula of 30!/4!(30-4)!, which leads me to 27405, but the key gives 810000, where I am a bit confused as how this number pops up as the answer

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2 Answers 2

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a) $30\cdot 29 \cdot 28 \cdot 27=657720$ (30 ways to choose a president, 29 ways to choose a vice president (one is taken for president) etc.)

b) $30\cdot 30 \cdot 30 \cdot 30=810000$

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  • $\begingroup$ thanks, this explains the situation. so here for every situation there will be 30 chances as each occurrence is not related to each other. $\endgroup$
    – Shariq
    Commented Sep 6, 2022 at 20:18
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    $\begingroup$ @Shariq , you're correct. $\endgroup$
    – eMathHelp
    Commented Sep 6, 2022 at 20:19
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Sketch for B:

Suppose one student, let's say his name is Donald, serves in all four posts. He is one of 30 students. Use the Fundamental Counting Principle, since we are allowing essentially drawing without replacement (multiply). This gives the solution.

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