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Math Help - Discrete Maths

  1. #1
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    Discrete Maths

    Hello again =)
    So lets see !

    1) We 've got a pouch with M&Ms candies . There are 4 Blue , 7 Orange , 8 Pink and 5 Yellow. Nick puts his hand into the pouch, and takes 6 candies.How many ways there exist , so that into the pouch there exist at least one Blue candy?

    I believe, that 1st we must find out how many different combinations of 6 candies can be done. I think its 24^6 = 191,102,976. Then i think that we have to find out how many combinations of 6 candies that include 4xBlues and 2 others can be done, and just subtract it.

    Basically i think the solution is 24^6 - 20^2 = 191,102,576


    2) How many 6-digit numbers can be made if:

    • The First digit must be 6,7 or 8
    • The last digit must be < = 5
    • Every digit must be unique

    Its 3x10x10x10x10x6 =180,000 . But, i think that, in this way, its not every digit unique. Any suggestions?

    Thanks in advance for any responses =)
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  2. #2
    Junior Member nimon's Avatar
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    n choose k

    1) For this question, the number of ways of picking 6 candies out of 24 is not 24^{6}, that would be the number different sequences you can get by picking six 6 but replacing each one as you put it back. The number of ways of choosing k candies from a bag of n can be shown to be
    \frac{n!}{(n-k)!k!}.
    You can find an explanation of this here: Combination - Wikipedia, the free encyclopedia
    Last edited by nimon; January 18th 2010 at 04:42 AM. Reason: Ahem... it said =< 5 not >= 5 !! :/
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  3. #3
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    Hello, primeimplicant!

    2) How many 6-digit numbers can be made if:

    . . The first digit must be 6, 7 or 8
    . . The last digit must be < 5
    . . Every digit must be unique

    The first digit is from: \{6, 7, 8\}\;\;\hdots\;\text{3 choices.}

    The last digit is from: \{0, 1, 2, 3, 4, 5\}\;\;\hdots\;\text{6 choices.}

    The middle 4 digits are selected from the remaining 8 digits . . . _8P_4 = 1680\text{ choices.}


    Answer: .  3 \times 6 \times 1680 \;=\;30,240

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  4. #4
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    Thanks, you make it soooo easy! Artist!
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  5. #5
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    Regarding my 1st question,
    with nimon's help, i believe its like this:



    \binom{24}{6}-\binom{20}{2}= 134,596 - 190= 134,406<br /> <br /> <br />

    So i believe 134,406 is the final answer , correct me if i 'am wrong =)
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  6. #6
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    Quote Originally Posted by primeimplicant View Post
    Regarding my 1st question,
    with nimon's help, i believe its like this:



    \binom{24}{6}-\binom{20}{2}= 134,596 - 190= 134,406<br /> <br /> <br />

    So i believe 134,406 is the final answer , correct me if i 'am wrong =)
    At least one blue: \binom{24}{6}-\binom{20}{6}
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  7. #7
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    \binom{24}{6}-\binom{20}{6}

    =(number\ of\ ways\ to\ choose\ any\ 6)-(number\ of\ ways\ to\ choose\ zero\ blues)

    is the number of ways to choose at least 1 blue.

    As this includes choosing 4 blues, there will be no blues left in the pouch for many of these selections.

    \binom{24}{6}-\binom{20}{2}

    is correct.

    \binom{24}{6}-\binom{20}{2}=\binom{4}{0}\binom{20}{6}+\binom{4}{  1}\binom{20}{5}+\binom{4}{2}\binom{20}{4}+\binom{4  }{3}\binom{20}{3}
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  8. #8
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    alright, i feel i little bit lost, and dizzy.

    Again , i want to find out how many ways there exist so that into the pouch, theres at least 1 Blue candy, Not into Nick's hand .

    So its

    \binom{24}{6}-\binom{20}{2}<br />
As i said before right?


    Edit: <br />
\binom{24}{6}-\binom{20}{2}<b>=\binom{4}{0}\binom{20}{6}+\binom{4}{1}\binom{20}{  5}+\binom{4}{2}\binom{20}{4}+\binom{4}{3}\binom{20  }{3}</b>

    I Dont understand the Bold part, i would i appreciate if you could link me some wiki related theory or something.

    Thanks again

    Edit2: ok nvm found it
    Last edited by primeimplicant; January 18th 2010 at 07:48 AM.
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  9. #9
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    That's it, primeimplicant,

    We have to distinguish between
    "at least one in the hand", or
    "at least one in the pouch".

    You're correct.
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  10. #10
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    \binom{24}{6}-\binom{4}{4}\binom{20}{2}=\binom{4}{0}\binom{20}{6  }+\binom{4}{1}\binom{20}{5}+\binom{4}{2}\binom{20}  {4}+\binom{4}{3}\binom{20}{3}

    This is

    (select any 6 from the 24)-(the number of ways to pair all 4 blues with 2 others from the remaining 20)

    and this will give us the number of ways to have at least 1 blue remaining in the pouch.

    This is the same as calculating the removal from the pouch of

    No blues with any 6 from the remaining 20
    + any blue from the 4 blues matched with 5 other non-blues
    + any 2 blues from the 4 with any 4 non-blues from the 20 non-blues
    + any 3 blues with any 3 non-blues

    as we must avoid taking out all 4 blues.
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