Chocolate Box Combinations
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.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
Note that at least one of the boxes must be a $ 1 $-piece box, because the number of chocolates ordered was odd. The problem is now to determine how many ways $ 14 $ pieces can be assembled using the $ 1 $, $ 2 $, and $ 4 $-piece boxes. If we begin with all $ 1 $-piece boxes, there's one way to do that. There are seven ways to have a mix of $ 1 $ and $ 2 $-piece boxes (one $ 2 $-piece, two $ 2 $-pieces, etc. all the way through seven $ 2 $-pieces). Now, each pair of $ 2 $-piece boxes can be replaced by a $ 4 $-piece box. If there is one $ 4 $-piece box, there are six ways to box the remaining ten pieces of chocolate with $ 1 $ and $ 2 $-piece boxes (no $ 2 $-pieces, one $ 2 $-piece, etc. all the way through five $ 2 $-pieces). If there are two $ 4 $-piece boxes, there are four ways to box the remaining six pieces of chocolate (zero through three $ 2 $-piece boxes). Finally, if there are three $ 4 $-piece boxes, there are two ways to box the remaining two pieces of chocolate (either no $ 2 $-piece boxes or one $ 2 $-piece box). Thus, there are a total of $ 1 + 7 + 6 + 4 + 2 = \boxed{20} $ combinations of boxes possible.
Freeflows