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    Justin1932's Avatar
    Justin1932 Posts: 2, Reputation: 1
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    #1

    Feb 24, 2010, 06:12 PM
    probability
    How many license plates of three letters and three digits are possible such that:

    a.) The letters and digits alternate? Note: The first position may be either a letter or a digit.

    b.) The letters are all grouped together?

    c.) There must be a letter at the beginning and the end of the plate?

    d.) The letters and digits may appear anywhere, with no restrictions on order?
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #2

    Feb 25, 2010, 02:59 AM

    a) I'll say the same thing here. I can see only DLDLDL and LDLDLD as possible combinations.

    b) Take all the letters as a whole entity. You now have one block of letters, with 3 digits, making a total of 4 terms. Use permutations/combinations to see how many combinations there are.

    c) Fix those two letters. You are left with one letter in the middle together with 3 digits. Permute the 4 inner terms to give the number of combinations you can have.

    d) That one is the easier compared to b) and c).

    Perhaps you don't quite understand how permutations and combinations work?

    If the plate had only two digits, one letter and one digit, the total number of combinations is given by: 26x10 = 260 possibilities. (there are 26 letters in the alphabet and 10 numbers including 0)
    For a three digit plate, given there are 2 letters and 1 digit, you have (26x26x10) = 6760 possibilities.
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #3

    Feb 25, 2010, 07:23 AM

    Let's look at the first question. There are two possible arrangements as Unk said: DLDLDL and LDLDLD. Each "D" can be one of 10 possible dgits, and each "L" is on of 26 possible letters, so the total number of license plates you can make is:

    10*26*10*26*10*26 + 26*10*26*10*26*10 = 2*(10^3)*(26^3) = 35,152,000 possible combinations.

    You can apply this same line of reasoning to the other 3 problems.

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