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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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!!!

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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I need the work with the answer please! @abd212
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!).

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Try answering this question if you can! @billj5
Try answering this question if you can! @Aureyliant
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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