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How many ways are there to pick a collection of 10 coins from piles of

  • #1
Hi all. I was hoping somebody could help me with my reasoning here? This is a twist on a typical balls and bins problem.

How many ways are there to pick a collection of 10 coins from piles of pennies, nickels, dimes, quarters, and half-dollars? Base on the following condition:

a) Assuming that each pile has at least 10 or more coins.
14choose4 options. So 1001 ways to pick a collection of 10 coins.

b) Assuming that each pile has at least 10 or more coins and the pick must consist of at least one quarter coin or at least one dime.
14choose4 - 12choose10 = 1001 - 66 = 935

c) There are only 8 coins in each pile.
11choose4 = 330

d) There are only 8 coins in each pile and the pick must have at least one penny and two nickels?
11choose4 - 1 = 329
 
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  • #2
haruspex
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I don't see a full statement of the question. If it was in the title, that has become truncated.
 
  • #3
I apologize. I'm new to the forums. I added the question to the body as well. Thanks
 
  • #4
haruspex
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a) Assuming that each pile has at least 10 or more coins.
14choose4 options. So 1001 ways to pick a collection of 10 coins.
Yes.
b) Assuming that each pile has at least 10 or more coins and the pick must consist of at least one quarter coin or at least one dime.
14choose4 - 12choose10 = 1001 - 66 = 935
Yes.
c) There are only 8 coins in each pile.
11choose4 = 330
I make it much more. What's your logic?
d) There are only 8 coins in each pile and the pick must have at least one penny and two nickels?
11choose4 - 1 = 329
Yes. (Doesn't this make it clear that 330 is not enough for c)?
 
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  • #5
Using a small script I wrote I see the answer for c should be 976, but I need to think about why that is. :)
 
  • #6
haruspex
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Using a small script I wrote I see the answer for c should be 976, but I need to think about why that is. :)
I would count how many of the 1001 you need to remove.
 
  • #7
Right, the answer there is 25. So the only thing I can really think of is that we have 5 bins and in each bin there's a maximum of 2 coins that can't be assigned to that bin, so we have to subtract 5^2 from 1001.

Is that on the right track?
 
  • #8
haruspex
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Right, the answer there is 25. So the only thing I can really think of is that we have 5 bins and in each bin there's a maximum of 2 coins that can't be assigned to that bin, so we have to subtract 5^2 from 1001.

Is that on the right track?
Not quite following it the way you express it. How about, the disallowed combinations involve choosing 9 from the same bin; there are 5 ways to pick that bin, and five ways to pick the 10th coin.
 
  • #9
That's a better way of explaining what I was trying to express :).

Thanks!
 
  • #10
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can anyone explain where these numbers are coming from? the 14choose4 , 12choose10, 11choose4??
 
  • #11
haruspex
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can anyone explain where these numbers are coming from? the 14choose4 , 12choose10, 11choose4??
I assumed cronuscronus was using the standard formula. Its derivation is quite interesting.
Suppose you have R identical objects to divide into N distinguished piles. One way to it physically is to lay he R objects out in a line and place N-1 dividers at different points in the line. So the objects before the first divider go into the first pile, and so on.
Now think of this a different way: you have R+N-1 things in a line, and you can choose which N-1 of them correspond to dividers, the remaining R corresponding to objects. With a little thought, you can see that there is a 1-1 relationship between the possibilities. So the number of ways of dividing the R objects into N piles is N+R-1CN-1, or equivalently N+R-1CR.
It's easy to extend this to R distinct objects, but there's no easy way to handle N non-distinct piles. That gets you into partition theory.
 

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