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# Counting with restrictions problem

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Hello! I was solving some problems when I came across this one and got stumped :( Help would be greatly appreciated!

A customer ordered 15 pieces of gourmet chocolate. The order can be packaged in small boxes that contain 1, 2 or 4 pieces of chocolate. Any box that is used must be full. How many different combinations of boxes can be used for the customer's 15 chocolate pieces? One such combination to be included is to use seven 2-piece boxes and one 1-piece box.

starsinajar  Mar 12, 2018
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A customer ordered 15 pieces of gourmet chocolate.
The order can be packaged in small boxes that contain 1, 2 or 4 pieces of chocolate.
Any box that is used must be full.
How many different combinations of boxes can be used for the customer's 15 chocolate pieces?
One such combination to be included is to use seven 2-piece boxes and one 1-piece box.

$$\small{\text{In 15 pieces max. \color{red}3 (\color{green}4-piece boxes), max. \color{red}7 (\color{green}2-piece boxes), and max. \color{red}15 (\color{green}1-piece boxes) }} \\\\$$

$$\small{ \begin{array}{|rcll|} \hline && \displaystyle \left(\sum\limits_{i=0}^{\color{red}3} x^{({\color{green}{4}}*i)} \right) \times \left(\sum\limits_{i=0}^{\color{red}7} x^{({\color{green}{2}}*i)} \right) \times \left(\sum\limits_{i=0}^{\color{red}15} x^{({\color{green}{1}}*i)} \right) \\\\ &=&(1+x^4+x^8+x^{12}) \times (1+x^2+x^4+x^6+x^8+x^{10}+x^{12}+x^{14}) \times \\ && \times (1+x^1+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^{10}+x^{11}+x^{12}+x^{13}+x^{14}+x^{15} ) \\\\ &=& x^{41} + x^{40} + 2 x^{39} + 2 x^{38} + 4 x^{37} + 4 x^{36} + 6 x^{35} + 6 x^{34} + 9 x^{33} + 9 x^{32} + 12 x^{31} + 12 x^{30} \\ &+& 16 x^{29} + 16 x^{28} + 20 x^{27} + 20 x^{26} + 22 x^{25} + 22 x^{24} + 24 x^{23} + 24 x^{22} + 24 x^{21} \\ &+& 24 x^{20} + 24 x^{19} + 24 x^{18} + 22 x^{17} + 22 x^{16} + {\color{red}20 x^{15}} + 20 x^{14} + 16 x^{13} + 16 x^{12} \\ &+& 12 x^{11} + 12 x^{10} + 9 x^9 + 9 x^8 + 6 x^7 + 6 x^6 + 4 x^5 + 4 x^4 + 2 x^3 + 2 x^2 + x + 1 \\ \hline \end{array} }$$

$$\text{The coefficient from \small{ \color[rgb]{1,0,0}{x^{15}} }  is \small{ \color[rgb]{1,0,0}{ 20 } } . } \\ \text{So there are \mathbf{20} possibilities.}$$

$$\begin{array}{|r|r|r|r|} \hline & \color{green}{1}\text{-piece boxes} & \color{green}{2}\text{-piece boxes} & \color{green}{4}\text{-piece boxes} \\ \hline 1 & 15 & & \\ \hline 2 & 13 & 1 & \\ \hline 3 & 11 & 2 & \\ \hline 4 & 11 & & 1 \\ \hline 5 & 9 & 1 & 1 \\ \hline 6 & 9 & 3 & \\ \hline 7 & 7 & & 2 \\ \hline 8 & 7 & 2 & 1 \\ \hline 9 & 7 & 4 & \\ \hline 10 & 5 & 1 & 2 \\ \hline 11 & 5 & 3 & 1 \\ \hline 12 & 5 & 5 & \\ \hline 13 & 3 & & 3 \\ \hline 14 & 3 & 2 & 2 \\ \hline 15 & 3 & 4 & 1 \\ \hline 16 & 3 & 6 & \\ \hline 17 & 1 & 1 & 3 \\ \hline 18 & 1 & 3 & 2 \\ \hline 19 & 1 & 5 & 1 \\ \hline 20 & 1 & 7 & \\ \hline \end{array}$$

heureka  Mar 12, 2018