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Math Help - How many combinations..?

  1. #1
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    How many combinations..?

    Hi!

    i am having this difficulty at the moment. My Dad has forgotten the combination to his bicycle lock... he knows (believes) it is a combination of 44670. We have tried the more obvious ones such as 07644, and moving the 0 around. I am wondering how many combinations of 44760 are there?

    if i've got nothing to do i i'll try them and see. What has stopped me from working it out is the 4 appearing twice, i don't know how that affects these kind of things.

    also, just as kind of "i want to know just so i don't try it" kind of thing, if the combination lock has 5 rows of ten numbers, how many combinations of those numbers can there be? we thought it was 10x10x10x10x10 = 100000

    Anyway, thanks for any help!

    James
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    Bar0n janvdl's Avatar
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    Quote Originally Posted by jiminwatford View Post
    Hi!

    i am having this difficulty at the moment. My Dad has forgotten the combination to his bicycle lock... he knows (believes) it is a combination of 44670. We have tried the more obvious ones such as 07644, and moving the 0 around. I am wondering how many combinations of 44760 are there?
    {5 \choose 2} = 10 (Assuming the there is one digit doubled, in this case the 4. Note the 4's need not be next to each other for this number of arrangements)

    (I suggest writing all the combinations down first on a piece of paper then trying them physically.)


    Quote Originally Posted by jiminwatford View Post
    if i've got nothing to do i i'll try them and see. What has stopped me from working it out is the 4 appearing twice, i don't know how that affects these kind of things.
    Are you quite certain the 4 appears twice? Maybe you should try doubling the 6 or the 7, etc in the combination instead of the 4.

    Quote Originally Posted by jiminwatford View Post
    also, just as kind of "i want to know just so i don't try it" kind of thing, if the combination lock has 5 rows of ten numbers, how many combinations of those numbers can there be? we thought it was 10x10x10x10x10 = 100000
    10^5 = 100 000 combinations.

    Quote Originally Posted by jiminwatford View Post
    Anyway, thanks for any help!

    James
    Hope I could help
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    hi, the four would be the only one doubled as he used the numbers from his birthday and had to use a 0 for the extra line.

    i'm not sure i follow. are you saying there is only 10 combinations those numbers can be arranged in? I must have misunderstood, as when i keep the 4s together there is 24 combinations

    thanks
    James
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    Bar0n janvdl's Avatar
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    Quote Originally Posted by jiminwatford View Post
    hi, the four would be the only one doubled as he used the numbers from his birthday and had to use a 0 for the extra line.

    i'm not sure i follow. are you saying there is only 10 combinations those numbers can be arranged in? I must have misunderstood, as when i keep the 4s together there is 24 combinations

    thanks
    James
    I may have made a mistake in my logic somewhere...

    Lets rather write it out...

    44670
    44760
    44607
    44706
    44076
    44067

    04467
    04476
    64470
    64407
    74460
    74406

    06447
    60447
    07446
    70446
    67440
    76440

    06744
    07644
    60744
    67044
    70644
    76044

    Which is 24 as you calculated. I'll rethink my calculations.

    ========

    What I would try:

    44076
    44706
    44607
    44067
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  5. #5
    Bar0n janvdl's Avatar
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    Oh i get it.

    4! gives us the answer because the fourth digit is the same as the fifth, giving us four distinct digits and not 5.
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  6. #6
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    Hello, James!

    My Dad has forgotten the combination to his bicycle lock.
    He believes it is a combination of 44670.
    I am wondering how many combinations of 44760 there are.

    Although it is called a "combination" lock, the order of the digits is significant.


    If those were five different digits, there would be 5! = 120 possible sequences.

    In each of the 120 sequences, the two 4's can be switched
    . . without creating a new sequence.

    Hence, our answer is twice as large as it should be.


    Therefore, the number of possible "combinations" is: . \frac{120}{2} \:=\:60


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Here's another approach . . .

    \text{We have five spaces to fill: }\;\_\;\;\_\;\;\_\;\;\_\;\;\_

    Pick two spaces for the two 4's . . . There are: . {5\choose2} \:=\:10 choices.

    The other three digits {0,6,7} can be placed in: . 3! \:=\:6 ways.

    Therefore, there are: .  10 \times 6 \:=\:60 possible "combinations".

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  7. #7
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, James!


    Although it is called a "combination" lock, the order of the digits is significant.


    If those were five different digits, there would be 5! = 120 possible sequences.

    In each of the 120 sequences, the two 4's can be switched
    . . without creating a new sequence.

    Hence, our answer is twice as large as it should be.


    Therefore, the number of possible sequences is: . \frac{120}{2} \:=\:60

    Soroban, if he says he knows for sure the two fours should be next to each other, wouldn't that change it? (Your solution assumes the fours need not be next to each other, am I right?)
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  8. #8
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    Hello, janvdl!

    Ha . . . I missed the additional constraint: adjacent 4's.


    In that case, we have four "numbers" to arrange: . \{0,\;\boxed{44},\:6,\:7\}

    . . Therefore, there are: . 4! = 24 "combinations".

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  9. #9
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    Hi, thanks!

    you are right, the 4s don't need to be together. i had worked out the 24 possibilities when keeping the 4s together and none of them worked, so i have to try them seperated. It is unlikely they are actually seperated as my Dad's birthday is in 19'44' so would probably have not seperated but... we need to try all things

    i just wish i hade the patience to try all 100,000 options!

    thanks
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  10. #10
    Bar0n janvdl's Avatar
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    Quote Originally Posted by jiminwatford View Post
    Hi, thanks!

    you are right, the 4s don't need to be together. i had worked out the 24 possibilities when keeping the 4s together and none of them worked, so i have to try them seperated. It is unlikely they are actually seperated as my Dad's birthday is in 19'44' so would probably have not seperated but... we need to try all things

    i just wish i hade the patience to try all 100,000 options!

    thanks
    Erm... Why not just grind the thing off? its much much easier, unless that lock is worth $1000
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