Alg2

what is the number of possibilities of arrangements of the letters in the word intercept?

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asked by Rock
  1. When there is no repetition of the letters, the number of permutations (order counts) is nPr, taking r letters at a time n!/(n-r)!.
    If all n letters are taken (as in the present case) to make the word, nPn reduces to
    n!/(n-n)!=n!/0!=n!/1 = n!

    However, if there are repetitions of letters, we have to reduce the number of permutations by the factor r1! for each occurrence of the duplication/repetition.

    Examples:
    1. How many arrangements of the six letters of the word "number" are there?
    6 letters, no repetitions, 6P6=6!=720.

    2. How many arrangements of the eleven letters of the word "mathematics" are there?
    We note repetitions of the letters m(2), a(2), t(2), so the number of arrangements
    = (11P11)/(2!2!2!)
    =11!/(2!2!2!)
    =4989600

    Post if you need more information.

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    posted by MathMate

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