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anonymous

  • one year ago

The number of ways in which we can get a score of 11 by throwing three dice is...

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  1. misty1212
    • one year ago
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    HI!!

  2. misty1212
    • one year ago
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    you familiar with dice?

  3. anonymous
    • one year ago
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    yeah i am..

  4. misty1212
    • one year ago
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    ok i guess we got a bunch of counting to do then because you have 3 dice total, not the usual 2

  5. anonymous
    • one year ago
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    yep two dices would be easy.. but there's 3

  6. misty1212
    • one year ago
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    none of the dice can show a one, but if one shows a 2, then it can be \[(2,5,6)\] or any combination of those there are 6 i believe \[(2,5,6),(2,6,5),(5,2,6),(5,6,2),(6,2,5),(6,5,2)\] or \(3!=6\) ways where one is a two

  7. anonymous
    • one year ago
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    18,27,45,56 are the options

  8. misty1212
    • one year ago
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    now lets move on to one showing a 3 \[(3,2,6)\] 6 of those \[(3,3,5)\] 3 of those \[(3,4,4)\] 3 of those guess that is it for one of them showing a 3

  9. misty1212
    • one year ago
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    ready to move on to 4? we have to make sure we don't count anything twice

  10. misty1212
    • one year ago
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    oh i made a mistake one could show a 1 lets count those too

  11. misty1212
    • one year ago
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    \[(1,5,5)\] 3 of those and \[(1,4,6)\] 6 of those

  12. misty1212
    • one year ago
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    keep counting in this way, you will get it

  13. misty1212
    • one year ago
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    oh nvm we are done just count them up

  14. anonymous
    • one year ago
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    you made 12 cases

  15. misty1212
    • one year ago
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    for two dice the pattern is \[1,2,3,4,5,6,5,4,3,2,1\]

  16. misty1212
    • one year ago
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    yeah count the number of ways for each for 3 dice the pattern is \[1,3,6,10,15,21,25,27,27,25,21,15,10,6,3,1\]

  17. misty1212
    • one year ago
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    for rolling respectively \[3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18\]

  18. misty1212
    • one year ago
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    notice that \[(1,4,6)\] where all the numbers are different has \(3!=6\) possible ways to do it

  19. anonymous
    • one year ago
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    its confusing in counting cases

  20. misty1212
    • one year ago
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    whereas \((1,5,5)\) has only 3 possible ways \[(1,5,5),(5,1,5),(5,5,1)\]

  21. misty1212
    • one year ago
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    ok lets go slow

  22. misty1212
    • one year ago
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    you got one possibility is \((1,4,6)\) right because they add up to 11

  23. misty1212
    • one year ago
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    but you have \(3!=6\) permutations of those three numbers

  24. misty1212
    • one year ago
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    6 is not so big that we cannot list them \[(1,4,6),(1,6,4),(4,1,6),(4,6,1),(6,1,4),(6,4,1)\] six ways to do it

  25. anonymous
    • one year ago
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    ok...

  26. misty1212
    • one year ago
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    so lets do them in order

  27. misty1212
    • one year ago
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    \[(1,4,6)\] 6 choices \[(1,5,5)\] 3 choices

  28. misty1212
    • one year ago
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    \[(2,3,6)\] 6 choices \[(2,4,5)\] 6 choices

  29. misty1212
    • one year ago
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    \[(3,3,5)\]3 choices \[(3,4,6)\] 6 choices

  30. triciaal
    • one year ago
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    |dw:1433100469127:dw|

  31. misty1212
    • one year ago
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    and that is all if you go further, you will get repeats

  32. misty1212
    • one year ago
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    add to get \[6+3+6+6+3+6=27\]

  33. triciaal
    • one year ago
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    my approach here is to find the possibilities of getting 11 when the first die is a 1 then when a 2 etc eventually will add the possibilities

  34. anonymous
    • one year ago
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    great @misty1212 thanks @triciaal

  35. misty1212
    • one year ago
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    \[\color\magenta\heartsuit\]

  36. triciaal
    • one year ago
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    |dw:1433101773240:dw|

  37. triciaal
    • one year ago
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    I see 2 errors P(2) = 4 and P(6) = 8

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