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Sat 12 Apr, 2008 01:09 pm
How many different combinations are there for substituting coins for a dollar, eg. 4 quarters; two quarters plus 50 pennies; ...?
There are five coins that can be used: penny, nickel, dime, quarter and fifty cent piece?
Is there a formula that one can use to figure this out?
Sure there's a formula, it's the old binomial coefficient:
http://mathworld.wolfram.com/Combination.html
Quote:The number of ways of picking k unordered outcomes from n possibilities. Also known as the binomial coefficient or choice number and read "n choose k"
_nC_k=(n; k)=(n!)/(k!(n-k)!)
Ahhh, the old binomial coefficient, how could I have forgotten that?
Thanks, HS, but I'm as lost as I was before. I'll look at the info at that site a bit more, but somehow I don't think it's gonna help.
If you want to know the number of ways to make change for a dollar, it's got nothing to do with binomial coefficients. Google "change for a dollar" and you'll find the answer.
markr wrote:If you want to know the number of ways to make change for a dollar, it's got nothing to do with binomial coefficients. Google "change for a dollar" and you'll find the answer.
Thanks, Markr. So, is there a formula for this type of math problem?
I reckon there would be about, lessee now,..... about 293 ways.
McTag wrote:I reckon there would be about, lessee now,..... about 293 ways.
You know what I've told you about showing your work, young man. If you don't show, your score is a no-go.
There isn't a formula, but there is a generating function. Go here for a nice discussion - it's near the bottom of the page.
http://www.cut-the-knot.org/ctk/GeneratingFunctions.shtml
Mac wrote-
Quote:I reckon there would be about, lessee now,..... about 293 ways.
In Zimbawe there's a lot more than that. I bet you could paint the town red with $10 there.
Not that I recommend trying it mind you.
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