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).
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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!).