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 Q51  V47
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Re: Permutation and Combination - Arranging Speakers with Conditions [#permalink]
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You could just imagine arranging the 8 people in a random order, and ask: what is the probability the CEO will come before the CTO, and the CTO before the COO? That probability is 1/6, because there are 6 orders the CEO, CTO and COO could end up speaking in, and each of those 6 orders is equally likely. So in 1/6 of all of the orders we could make, the CEO, CTO and COO will end up speaking in that order, and the answer is 8!/6.
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Re: Permutation and Combination - Arranging Speakers with Conditions [#permalink]
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And that's the same answer as your friend got -- your friend might have thought of it this way: "we have 8 slots on the speaking schedule to fill, and we can choose three of them for the CEO, CTO and COO in 8C3 ways. Once we choose those slots the sequence is predetermined: the CEO goes in the first slot we choose, the CTO in the next, the COO in the last. Then there are 5 slots left to fill using the other five people, and we have 5 choices for the first, 4 for the next, and so on, and the answer is 8C3 * 5!." That also gives the answer.
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Permutation and Combination - Arranging Speakers with Conditions [#permalink]
Ian, your answer makes sense! Thanks for sharing a new insight - combining probability and combinations to figure out the problem!
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Re: Permutation and Combination - Arranging Speakers with Conditions [#permalink]
8C3 * 5! is the correct answer..Thanks,,
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Re: Permutation and Combination - Arranging Speakers with Conditions [#permalink]
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