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anonymous

  • 4 years ago

I have some questions regarding combinations in finite. Here is my first one I am stumped on: Select 3 coins, but assume that there are 5 dimes, 3 nickels, and 2 quarters. In How many possible ways can the selection be made so that the value of the coins is AT LEAST 25 cents?_______________________________ Any help? Formulas have not done me so well

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  1. anonymous
    • 4 years ago
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    i think you have to grind it til you find it

  2. anonymous
    • 4 years ago
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    sorry i was away

  3. anonymous
    • 4 years ago
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    oh good so i am less embarrassed. you have to work in cases

  4. anonymous
    • 4 years ago
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    i have tried a number of times but cant solve it

  5. anonymous
    • 4 years ago
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    cant break down the cases correctly

  6. anonymous
    • 4 years ago
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    i assume

  7. anonymous
    • 4 years ago
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    1) two quarters one dime only one way to select two quarters, 5 choices for the dime, so 5 in that case 2) two quarters one nickel, 3 choices for the nickel so 3 ways in that case

  8. anonymous
    • 4 years ago
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    3) one quarter two dimes \[2\times \dbinom{5}{2}=2\times 10=20\]

  9. anonymous
    • 4 years ago
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    4) one quarter two nickels \[2\times 3=6\] 5) one quarter one dime one nickel \[2\times 3\times 5=30\]

  10. anonymous
    • 4 years ago
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    and finally two dimes and a nickel \[10\times 3=30\] think that is all, so if we add them up we should get the right answer

  11. anonymous
    • 4 years ago
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    @zarkon, you got a snap way to do this, or it it just grind it out?

  12. anonymous
    • 4 years ago
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    wow i did not break the cases down enough I will check and see if it is right

  13. anonymous
    • 4 years ago
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    damn i forgot 3 dimes!!

  14. Zarkon
    • 4 years ago
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    I do have a quicker way...but it is goog to do it this way...you also nee 3 dimes I believe

  15. anonymous
    • 4 years ago
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    I will try to add them up and see

  16. anonymous
    • 4 years ago
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    add another 10 what is the quick way for my benefit

  17. Zarkon
    • 4 years ago
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    which gives a total of 104

  18. Zarkon
    • 4 years ago
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    you can also look at it this way...

  19. anonymous
    • 4 years ago
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    now i get to learn something

  20. Zarkon
    • 4 years ago
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    there are \[_{10}C_{3}=120\] ways to pick the 3 coins... only NNN and 2nickles and a dime are less than 25 cents 120-1-15=104

  21. anonymous
    • 4 years ago
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    yes much snappier

  22. Zarkon
    • 4 years ago
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    I don't see a way to do it without some cases

  23. anonymous
    • 4 years ago
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    these few are killing me

  24. anonymous
    • 4 years ago
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    104 is correct

  25. anonymous
    • 4 years ago
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    yes well that would really be magic

  26. anonymous
    • 4 years ago
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    thank you so much I have some more that I am stumped as well if you want to crack them

  27. anonymous
    • 4 years ago
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    ?:)

  28. anonymous
    • 4 years ago
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    post a new thread, you will get lots of answers i am sure

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