anonymous
  • anonymous
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?______________
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
(6C4)+(7C1)
anonymous
  • anonymous
6x5x4x3 -------- = 15 +7= 22 4x3x2x1 Is that right?
anonymous
  • anonymous
that didnt work

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anonymous
  • anonymous
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
Directrix
  • Directrix
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.
Directrix
  • Directrix
I included the case of 6 dimes and messed up the first time.
anonymous
  • anonymous
That did it direct.... awesome man
anonymous
  • anonymous
it is hard for me to id the cases in a lot of these
Directrix
  • Directrix
It is tricky.
anonymous
  • anonymous
1 more i couldn't do from that section i am going to post.... thank you so much
Directrix
  • Directrix
Ok, I'll look for it.

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