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    i'm stuck on a question please help me>>>

    a) A code for a mobile phone is made up of three digits. The digits use the numbers 0-9. The digits can be repeated. How many different possible codes are there?

    b) If the digits cannot be repeated, how many different codes are there?
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    use the the rule of permutation and combination
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    (Original post by channie11223)
    i'm stuck on a question please help me>>>

    a) A code for a mobile phone is made up of three digits. The digits use the numbers 0-9. The digits can be repeated. How many different possible codes are there?

    b) If the digits cannot be repeated, how many different codes are there?
    Imagine there are three gaps which you need to fill with the digits:

    _ _ _

    How many choices for the first digit? Second digit? Third digit? Then use the product rule for counting (multiply the choices) to get the answer.
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    a) Think of the code as three different boxes, which we will place single digits inside.

    Think of the amount of options there are for each box, remembering that there are ten digits from 0-9.

    Then use the rule of product to work out the total amount of codes,

    b) Think about how not being able to repeat digits affects the amount of options we have for the second and third boxes.
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    (Original post by aryanmalik567)
    x
    Hi, please edit your post to remove the full solution and just give hints.

    Posting full solutions is against the guidelines of the maths forum.
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    (Original post by Notnek)
    Hi, please edit your post to remove the full solution and just give hints.

    Posting full solutions is against the guidelines of the maths forum.
    I have edited, I apologise I was not aware of the guidelines.
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    (Original post by aryanmalik567)
    I have edited, I apologise I was not aware of the guidelines.
    Don't worry, most new posters aren't. Please continue to help students in the maths forum
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    (Original post by aryanmalik567)
    a) Think of the code as three different boxes, which we will place single digits inside.

    Think of the amount of options there are for each box, remembering that there are ten digits from 0-9.

    Then use the rule of product to work out the total amount of codes,

    b) Think about how not being able to repeat digits affects the amount of options we have for the second and third boxes.
    Thanks for your help but i still don't understand, the repeating digits thing is confusing... I think a is 10x10x10=1000 but i'm not sure about b.
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    The code uses the digits 0-9, therefore there are 10 digits that can be used for each digit in the code, therefore the number of combinations when any number can be used up to 3 times is just 10*10*10 = 10^3 = 1000, this is your answer to part a. For part B, the number of possible choices each time decreases by 1 as the number is used, to calculate this you use the formula nPr (it should be on your calculator) where n is the number of digits that can be used (0-9 so 10) and r is the number of digits being chosen, this gives the number of permutations of 3 digit long numbers using that number of digits so you get 10P3 = 720 permutations of these numbers. Note that permutations are different from combinations in that permutations accounts for order, i.e. the codes 123 and 321 are counted seperately despite using the same digits whereas combinations would only count this once as the numbers used are the same
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    (Original post by channie11223)
    Thanks for your help but i still don't understand, the repeating digits thing is confusing... I think a is 10x10x10=1000 but i'm not sure about b.
    Think about this, if there are 10 counters, and we go to pick three without replacement, we have 10 counters to choose from for the first pick. How many counters are remaining for the second pick, and as a result, how many are available to choose from for the third pick?
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    (Original post by MattSull)
    The code uses the digits 0-9, therefore there are 10 digits that can be used for each digit in the code, therefore the number of combinations when any number can be used up to 3 times is just 10*10*10 = 10^3 = 1000, this is your answer to part a. For part B, the number of possible choices each time decreases by 1 as the number is used, to calculate this you use the formula nPr (it should be on your calculator) where n is the number of digits that can be used (0-9 so 10) and r is the number of digits being chosen, this gives the number of permutations of 3 digit long numbers using that number of digits so you get 10P3 = 720 permutations of these numbers. Note that permutations are different from combinations in that permutations accounts for order, i.e. the codes 123 and 321 are counted seperately despite using the same digits whereas combinations would only count this once as the numbers used are the same
    oh I see thank you!
 
 
 
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