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anonymous

  • one year ago

Donald has nine one-day passes to Dizzyworld. He can go alone and use one, or he can take a friend and use two. If he visits every day until he uses all the passes, in how many different ways can he use them? Using two a day for four days and then going alone at the end (2, 2, 2, 2, 1) is different from reversing the order (1, 2, 2, 2, 2). I NEED THE ANSWER WITH THE WORK TO THIS QUESTION!!!

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  1. anonymous
    • one year ago
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    55

  2. anonymous
    • one year ago
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    I need the work with the answer please! @abd212

  3. anonymous
    • one year ago
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    Case I: He goes by himself for 9 days: There is only 1 way in this case. Case II: He goes by himself for 7 days: There's 8!/7! ways this is possible. Case III: He goes by himself for 5 days: It's something like (1, 1, 1, 1, 1, 2, 2) There are 7!/(2!*5!) no of ways. Case IV: He goes by himself for 6 days: It's something like (1, 1, 1, 2, 2, 2) There are 6!/(3!*3!) ways this is possible. Make all the cases possible. Sum the number of ways and you will have your answer. In case you are wondering what formula I am using then its quite simple. If you have n objects and you have say r of one kind and s of another kind then, the total possible ways of arranging them is n!/(r!*s!).

  4. anonymous
    • one year ago
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    Try answering this question if you can! @billj5

  5. anonymous
    • one year ago
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    Try answering this question if you can! @Aureyliant

  6. anonymous
    • one year ago
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    he meant 3 days where he said 6, and he didnt include the case where he goes by himself for 1 day but other than that it looks good

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