in how many ways can G,H,I,I, and J be arranged to compose a 5-letter code so that two "I" are not next to each other?
OA[spoiler]=60-24=36[/spoiler]
OA[spoiler]=60-24=36[/spoiler]
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Well total ways of arranging 5 letters with 2 common letters=5!/2!=60 and 4!=24DanaJ wrote:Total ways of arranging the 5 letters will be 5! = 60.
Then you need to subtract the cases where you have the two I's next to each other. In such cases, they'd create one single, indivisible group. So you can safely say that the number of cases in which you have II is the number of possible arrangements of (G), (H), (II) and (J) or possible arrangements of 4 objects, i.e. 4! = 24.
You're left with the answer you provided.
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