Marie has five television shows recorded on DVR. She has enough time to watch three of them today and must decide in what order she will watch them. How many different orderings of the three shows she watches today can Marie choose?
My answer is 10.
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How did you get ten? – Thomas Andrews Dec 03 '19 at 03:22
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I used the combination formula. Is it correct? – Ocean Dec 03 '19 at 03:23
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1Combinations don’t count the ordering. There are $\binom5 3=10$ ways to pick three shows, but that doesn’t order them. – Thomas Andrews Dec 03 '19 at 03:24
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1Suppose the five shows are A, B, C, D and E, and Marie chooses to whatch A, B and C. She can watch those three in order ABC, ACB, BAC, BCA, CAB, CBA. But if she chooses to watch C, D and E, she can watch those in the order CDE, CED, DCE, DEC, ECD and EDC. That's already 12. So you need to account not only for which shows she chooses to watch out of the five, but also in what order. – Randy Marsh Dec 03 '19 at 03:25
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Is my answer correct or not? I’m confused – Ocean Dec 03 '19 at 03:27
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Aren’t you supposed to use the combinations formula??? – Ocean Dec 03 '19 at 03:28
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Pick the first show. There are $5$ options.
Pick the second show -- there are $4$ options left.
Pick the third show -- now you only have $3$ options.
$5 \times 4 \times 3 = 60$ orders.
Saketh Malyala
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I solved the problem as a combination. I see what i did. I’m supposed to order it so it’ll be a permutation. Thanks so much. – Ocean Dec 03 '19 at 03:31
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I get confused sometimes. How would you know if a problem is a permutation or combination? – Ocean Dec 03 '19 at 03:35
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if order doesn't matter, then use a combination. for example, how many ways can i pick 4 of my 10 friends to take to a movie, i'd use a combination, because it doesn't matter what order. if i had to line up 4 of my 10 gifts for a picture, then i'd use a permutation, because order matters – Saketh Malyala Dec 03 '19 at 03:37