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    4. how many combinations are there for a - z and 0 - 9
    how many combinations are there for a - z and 0 - 9

    in order to issue license plates.

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    0  Views: 421 Answers: 1 Posted: 13 years ago
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      1 Answer

      26P3
      nPr=n! /(n-r)!
      26! / 23!
      where n!=n(n-1)(n-2).....(1)
      26 x 25 x 24=15,600
      ABC and BCA are different here. You should have meant permuations.

      13 years ago. Rating: 2
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