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mathmath333

  • one year ago

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{Their are 4 apples, 5 mangoes and 6 watermelons .}\hspace{.33em}\\~\\ & \normalsize \text{Find the number of ways }\hspace{.33em}\\~\\ & \normalsize \text{i.) I can purchase at least each of them. }\hspace{.33em}\\~\\ & \normalsize \text{ii.) I can purchase at least one of them. }\hspace{.33em}\\~\\ \end{align}}\)

  2. ganeshie8
    • one year ago
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    notice that apples have 5 states : {0, 1, 2, 3, 4} mangoes have 6 states : {0, 1, 2, 3, 4, 5} watermelons have 7 states : {0, 1, 2, 3, 4, 5, 6}

  3. mathmath333
    • one year ago
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    yes

  4. mathmath333
    • one year ago
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    is this correct =(2^4-1)*(2^5-1)*(2^6-1)

  5. anonymous
    • one year ago
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    I believe so

  6. anonymous
    • one year ago
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    the english is a bit unclear to me " I can purchase at least each of them"

  7. mathmath333
    • one year ago
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    but it is given 4*5*6 ways

  8. mathmath333
    • one year ago
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    in book

  9. anonymous
    • one year ago
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    thats the answer for part a) ?

  10. mathmath333
    • one year ago
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    yes

  11. ganeshie8
    • one year ago
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    for partb, is the answer 5*6*7 - 1 ?

  12. mathmath333
    • one year ago
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    yes for part b

  13. ganeshie8
    • one year ago
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    it seems textbook is treating all the fruit types as indistinguishable @jayzdd

  14. anonymous
    • one year ago
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    yes , hmm i guess i treated them as distinguishable

  15. mathmath333
    • one year ago
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    all apples are are identical and so all mango

  16. anonymous
    • one year ago
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    For the apple choice you can pick 1 apple 2 apples 3 apples 4 apples a total of 4 ways to select the apples Then multiply this by the number of selections you can make for mango 1 mango 2 mangoes 3 mangoes 4 mongoes 5 mangoes a total of 5 choices and then similarly for the watermelon there are 6 ways to select them by multiplication principle, 4*5*6

  17. mathmath333
    • one year ago
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    ok

  18. anonymous
    • one year ago
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    for part b) now include the zero case

  19. anonymous
    • one year ago
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    For the apple choice you can pick 0 apples, 1 apple, 2 apples, 3 apples , 4 apples a total of 5 ways to select the apples Then multiply this by the number of selections you can make for mango 0 mango 1 mango 2 mangoes 3 mangoes 4 mangoes 5 mangoes a total of 6 ways to select And then similarly for the watermelon 0 watermelon, 1 watermelon, ... 6 watermelon there are 7 ways to select them By multiplication principle, 5*6*7 But the directions say at least one fruit, so we cannot have 0 apple and 0 mangoe, and 0 watermelon. so we subtract 1

  20. mathmath333
    • one year ago
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    like these method in the previuos question there were 5 methods for bulb so how can i judge distinguishable bulbs or fruits

  21. anonymous
    • one year ago
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    The bulbs were fixed by the socket, so there was an ordering to them. lightbulb 1, lightbulb2, ... lightbulb 5 Here there is no natural order to the fruits, since they are indistinguishable. If they want to treat the fruits as distinguishable, the directions would explicitly say it. otherwise assume indistinguishable

  22. anonymous
    • one year ago
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    |dw:1440349712575:dw|

  23. anonymous
    • one year ago
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    the sockets are distinguishable i would say , and there are 2 states for each socket (on or off)

  24. anonymous
    • one year ago
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    |dw:1440349885112:dw|

  25. anonymous
    • one year ago
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    If the lightbulbs had no sockets, then the lightbulbs would be indistinguishable. But then there would be no light either. Since lightbulbs had light, we could assume they were connected to distinguishable sockets, and cannot move. This forces an ordering to the lightbulbs that make them distinguishable.

  26. anonymous
    • one year ago
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    The lightbulbs connected to the sockets are fixed in space and bolted down, because they are constructed that way by the carpenter. But the apples and mangoe fruits are not bolted down, so what matters is how many you pick, not which one

  27. anonymous
    • one year ago
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    Think of it this way. When you select the fruits, imagine you are blindfolded. There are three bags, a bag for apples , a bag for mangoes, bag for watermelons. You are instructed to select as many fruits from each bag as you want. The apples are all the same (let's imagine for the sake of argument). The mangoes are all exactly the same and the watermelons the same. you can select 0 apples, 1 apple, 2 apples... up to 4 apples. you can select 0 mangoes up to 5 mangoes, etc Doing this blindfold experiment would not make sense for the lighbulb example.

  28. mathmath333
    • one year ago
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    sry was afk

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