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anonymous

  • 5 years ago

how many ways can limitabilty be arranged if the first letter must be an I and the last letter must not be a Y?

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  1. sid1729
    • 5 years ago
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    the number of letters in "Limitabilty" is 11 (are you sure you did not mis-spell it?) Since the first letter has to be an L, number of open spots left is 10. Now, let's take Y. The number of possible spots Y can go into is 9 because the first spot is taken by L, and the last must not be y. Let Y take one of the 9 spots. Next, take the first "i". It can be placed into 9 spots again, because the first one is fixed, and Y took a spot. Similarly, the 2nd "i" can take 8 places, the third "i" can take 7 places. "m" can take 6, the first "t" can take 5, "a" can take 4 spots, "b" can take 3, the second "L" can take 2, and the second "t" can take 1. It is not important which order you do this in, the answer should be the same because we are calculating the number of spots each letter can take, keeping into account the spots taken by the previous letter we chose. so, the answer is 9 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

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