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anonymous

  • one year ago

MEDAL FAN AND TESTIMONIAL:on a camping trip you bring 12 items for 4 dinners for each dinner you use 3 items in how many ways can you choose the 3 items for the first dinner? for the second? for the third? for the fourth?

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  1. anonymous
    • one year ago
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    its on permutations and combinations algebra 2

  2. anonymous
    • one year ago
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    @Preetha

  3. anonymous
    • one year ago
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    @mathstudent55 ???

  4. anonymous
    • one year ago
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    It is combinations. And the result is same for the all four dinners.

  5. anonymous
    • one year ago
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    how do i do it?

  6. anonymous
    • one year ago
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    I can do the problems that are already set up but not word problems where i have to set it up

  7. anonymous
    • one year ago
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    brb

  8. anonymous
    • one year ago
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    You have four dinner. And for all of them, you need to make a decision to pick 3 items among 12 items. So 3 combinations among 12 item is: \[\left(\begin{matrix}12 \\ 3\end{matrix}\right) = \frac{ 12 \times 11 \times 10 }{ 9! \times 3! }\] I had a mistake though, in my first statement, where I said that they were equal. It is not. If you pick 3 items among 12, you will have 9 items left, since you took 3 of them.

  9. anonymous
    • one year ago
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    eek i got a weird answer

  10. anonymous
    • one year ago
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    what did you get?

  11. anonymous
    • one year ago
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    \[11/18144\]

  12. anonymous
    • one year ago
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    that was evaluating the right side

  13. anonymous
    • one year ago
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    ohh sorry, I forgot to add factorial sign to the 12. It is like this: \[\frac{ 12! }{ 9! \times 3! }\] it is originally this way.

  14. anonymous
    • one year ago
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    The general formula is: \[\left(\begin{matrix}c \\ n\end{matrix}\right) = \frac{ c! }{ n! \times (c - n)! }\]

  15. anonymous
    • one year ago
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    i got 220

  16. anonymous
    • one year ago
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    is that for the first dinner

  17. anonymous
    • one year ago
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    yes

  18. anonymous
    • one year ago
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    ok how do i find the others

  19. anonymous
    • one year ago
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    For the second dinner you will calculate this: \[\left(\begin{matrix}9 \\ 3\end{matrix}\right)\] Since we took the first three items, 9 left in the item list.

  20. anonymous
    • one year ago
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    not sure how to solve that?

  21. anonymous
    • one year ago
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    its 9/3??

  22. anonymous
    • one year ago
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    Check my answer, I generalized the formula for you

  23. anonymous
    • one year ago
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    \[\left(\begin{matrix}c \\ n\end{matrix}\right) = \frac{c!}{n! \times (c-n)!}\]

  24. anonymous
    • one year ago
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    i got 84

  25. anonymous
    • one year ago
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    yeah, probably, I didn't calculated

  26. anonymous
    • one year ago
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    ok 3rd?

  27. anonymous
    • one year ago
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    How many items do we left? We used 6 of them so far right?

  28. anonymous
    • one year ago
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    6

  29. anonymous
    • one year ago
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    yes, you will chose 3 from 6. Use the formula that I gave you. Did you understand how to use the formula?

  30. anonymous
    • one year ago
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    mm let me see ill try

  31. anonymous
    • one year ago
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    i got 20

  32. anonymous
    • one year ago
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    yes. I think

  33. anonymous
    • one year ago
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    ok for the fourth then all of the variables in the formula you gave are 3 right

  34. anonymous
    • one year ago
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    yes, you have learned ;)

  35. anonymous
    • one year ago
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    i think so but the problem is that I'm getting 1/6?

  36. anonymous
    • one year ago
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    you should get 1. Check if you didn't make any calculation mistakes.

  37. anonymous
    • one year ago
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    or i guess i would just say one because there isn't anything left

  38. anonymous
    • one year ago
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    yes :)

  39. anonymous
    • one year ago
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    oh haha okay thank you so much (:

  40. anonymous
    • one year ago
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    Anytime ;)

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