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anonymous

  • 4 years ago

Assume you have 6 dimes and 5 quarters (all distinct), and you select 5 coins. Total ways selection can be made= 462 How many ways can the selection be made so that at least 4 coins are dimes?______________

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  1. anonymous
    • 4 years ago
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    (6C4)+(7C1)

  2. anonymous
    • 4 years ago
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    6x5x4x3 -------- = 15 +7= 22 4x3x2x1 Is that right?

  3. anonymous
    • 4 years ago
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    that didnt work

  4. anonymous
    • 4 years ago
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    6*5*4*3*2*1 7*6*5*4*3*2*1 ------------ + -------------- 4*3*2*1*2*1 1*6*5*4*3*2*1

  5. Directrix
    • 4 years ago
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    Two separate cases: That would be the sum of the numbers of ways to choose exactly 4 dimes and then exactly 5 dimes.. C(6,4) times C(5,1) + C(6,5) times C(5, 0) = 81.

  6. Directrix
    • 4 years ago
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    I included the case of 6 dimes and messed up the first time.

  7. anonymous
    • 4 years ago
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    That did it direct.... awesome man

  8. anonymous
    • 4 years ago
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    it is hard for me to id the cases in a lot of these

  9. Directrix
    • 4 years ago
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    It is tricky.

  10. anonymous
    • 4 years ago
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    1 more i couldn't do from that section i am going to post.... thank you so much

  11. Directrix
    • 4 years ago
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    Ok, I'll look for it.

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