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if a license plate consists of 2 letters followed by 4 digits, how many different plates could be created having at least one letter or digit repeated?

I know its not 6,760,000

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  1. consider the Pr of two letters the same: 1/26

    Now consider the Pr of two numbers: 1/10*9/10*8/10
    Now the Pr of three numbers the same:
    1/10*1/10*9/10
    now the Pr of four numbers the same:
    1/10*1/10*1/10
    now the pr of having two pairs of numbers the same:
    1/10*9/10*1/9

    Pr(at least one letter or digit repeated:
    sum of (two letters, no numbers) + (two letters, two numbers)+(two letters, three numbers)+(two letters, four numbers)+(two letters, two pairs of numbers) + (no letters, two numbers)+(no letters, threenumbers)+(no letters, four numbers)+ (no letters,two pairs of numbers)

    I will do the first four probabilities for you:
    Pr(twoletters, no numbers)=1/26*9/10*8/10*7/10
    Pr(two letters, two numbers)=1/26*1/10*9/10*8/10
    Pr(two letters, three numbers)=1/26*1/10*1/10*9/10
    See if you can finish.

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    bobpursley
  2. Your asking for how many combinations and not the probability so ignore the answer above.

    First find the number of combinations if all letters and numbers can be repeated
    (26*26*10*10*10*10)=6760000

    Then find the number of combinations if none were repeated
    (26*25*10*9*8*7)=3276000

    (value of combinations that can be repeated)-(value of combinations that no letter or number could be repeated) = 6760000-3276000= 3484000

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