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Let $A = \{1,2,3,4,5,6\}$.

In how many ways can we select in order without replacement, three elements from A such that the last number is even:

The dr's solution was $(5 * 4) * 3 = 5P2 * 3 = 60$. But that doesn't make sense to me. Can someone please explain how we solved this?

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  • $\begingroup$ No idea. However, I know that the answer is $~\dfrac{6!}{3!} \times \dfrac{1}{2}.~$ This follows because, by symmetry, the last number chosen is just as likely to be even as odd. $\endgroup$
    – user2661923
    Mar 28 at 7:50

3 Answers 3

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Pick the even number first, and then pick two from the rest.

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I think the question is equivalent for the case when any one number must be even and not specifically the last one, if we choose the even one first is the same because that would only change the solution to read the sequence backwards. Therefore we have 3 options to choose from first, than 5, than 4.

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  • $\begingroup$ I thought so as well, but then I thought that if the last is even then the first 2 might be even or odd like this (OOE, EOE, OEE, EEE) and I don't know how to account for that in my calculation. Maybe I'm over-complicating things or the question isn't worded clearly enough. $\endgroup$
    – Ahmad Masoora
    Mar 28 at 8:00
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    $\begingroup$ But when you choose the first to be mandatorily even and read the result backwards, then the last one will be definitely even. And this method doesn't change the number of outputs. $\endgroup$
    – Dávid Laczkó
    Mar 28 at 8:10
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There are $\color{red}3$ possibilities for the last number. After we chose the last number, $5$ numbers remain. Now, there are $\color{red}5$ possibilities for choosing the first number. Then, $\color{red}4$ numbers remain to choose the second number. Overall, the number of the possibilities are $\color{red}{3\times 5\times 4=60.}$

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  • $\begingroup$ I am just showing my thinking about "this question". $\endgroup$
    – Bob Dobbs
    Mar 28 at 9:38

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