+26393 PSL4#18
\(\begin{array}{|c|c|c|c|c|} \hline & \text{penny} & \text{nickel} & \text{dime}& \text{quater} \\ & 1\ \text{cent} & 5\ \text{cent} & 10\ \text{cent}& 25\ \text{cent} \\ \hline &0&0&0&3 \\ &0&0&3&0 \\ &0&3&0&0 \\ &3&0&0&0 \\ \hline &1&2&0&0 \\ &1&0&2&0 \\ &1&0&0&2 \\ &0&1&2&0 \\ &0&1&0&2 \\ &0&0&1&2 \\ \hline &2&1&0&0 \\ &2&0&1&0 \\ &2&0&0&1 \\ &0&2&1&0 \\ &0&2&0&1 \\ &0&0&2&1 \\ \hline &1&1&1&0 \\ &1&1&0&1 \\ &1&0&1&1 \\ &0&1&1&1 \\ \hline sum &15&15&15&15 \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \text{sum} &=& 15\times 1\ \text{cent}+ 15\times 5\ \text{cent} + 15\times 10\ \text{cent}+ 15\times 25\ \text{cent} \\ &=& 15 + 75 + 150 + 375 \\ &=& 615 \\ &=& \mathbf{$6.15} \\ \hline \end{array}\)
+6251 \(\text{The first term comes from the stars and bars solution to the number of ways to allot coins into bins.$~\\$ In this case we allot 3 coins into 4 bins labeled pennies, nickels, dimes, quarters.}\\~\\ \text{The term $\dfrac{0.01+0.05+0.10+0.25}{4}$ represents the average value of a coin}\\~\\ \text{There are then 3 coins so the value is 3 times the average coin value}\\~\\ \text{Put all that together and you get the very clever formula posted}\)
btw I'm not the original poster of the solution, but who knows if they'll be back and it warrented explanation.
