In how many ways can I have $1.50 using exactly 50 Coins? The coins may be pennies (1 cent), nickels (5 cents), dimes (10 cents) or quarters (25 cents).
In how many ways can I have $1.50 using exactly 50 Coins?
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combinatorics
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0There is a solution of this problem, by a method quick and simple enough to do by hand, in the "Generating Functions" chapter of *Concrete Mathematics* by Graham, Knuth, and Patashnik. It is different from the three solutions below. – 2012-10-20
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2In how many ways can this problem be answered? – 2012-10-20
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3Moreover, each and every way will elicit a scowl on the cashier's face when you try to use them to purchase a candy bar! – 2012-10-20
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0As a comparison, if we differentiate arrangements of the coins, there are a lot more ways of arranging the change. Each choice of $x$, $x^5$, $x^{10}$, or $x^{25}$ when expanding the product $$ (x+x^5+x^{10}+x^{25})^{50}=\dots+277961174530259752\,x^{150}+\dots $$ corresponds to a particular choice of each of the $50$ coins. The coefficient given above yields the number that total to $\$1.50$. – 2012-10-20