# A customer ordered fifteen Zingers. Zingers are placed in packages of four, three, or one. In how many different ways can this order be filled?

Feb 12, 2017

$15$

#### Explanation:

There are at most $3$ packages of four Zingers. Treating each possibility $3 , 2 , 1 , 0$ as a separate case, there are then subcases according to the number of packages of three Zingers.

It's probably best to just systematically enumerate them as follows:

$o o o o \textcolor{w h i t e}{o} o o o o \textcolor{w h i t e}{o} o o o o \textcolor{w h i t e}{o} o o o$
$o o o o \textcolor{w h i t e}{o} o o o o \textcolor{w h i t e}{o} o o o o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o$

$o o o o \textcolor{w h i t e}{o} o o o o \textcolor{w h i t e}{o} o o o \textcolor{w h i t e}{o} o o o \textcolor{w h i t e}{o} o$
$o o o o \textcolor{w h i t e}{o} o o o o \textcolor{w h i t e}{o} o o o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o$
$o o o o \textcolor{w h i t e}{o} o o o o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o$

$o o o o \textcolor{w h i t e}{o} o o o \textcolor{w h i t e}{o} o o o \textcolor{w h i t e}{o} o o o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o$
$o o o o \textcolor{w h i t e}{o} o o o \textcolor{w h i t e}{o} o o o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o$
$o o o o \textcolor{w h i t e}{o} o o o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o$
$o o o o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o$

$o o o \textcolor{w h i t e}{o} o o o \textcolor{w h i t e}{o} o o o \textcolor{w h i t e}{o} o o o \textcolor{w h i t e}{o} o o o$
$o o o \textcolor{w h i t e}{o} o o o \textcolor{w h i t e}{o} o o o \textcolor{w h i t e}{o} o o o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o$
$o o o \textcolor{w h i t e}{o} o o o \textcolor{w h i t e}{o} o o o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o$
$o o o \textcolor{w h i t e}{o} o o o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o$
$o o o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o$
$o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o \textcolor{w h i t e}{o} o$

I count $15$ arrangements.

Feb 12, 2017

See below.

#### Explanation:

Calling $\left\{\alpha , \beta , \gamma\right\} \in \mathbb{Z} \ge 0$ we have

$3 \alpha + 2 \beta + \gamma = 15$

The different solutions to this equation, calling diophantine equation after Diophantus of Alexandria, equals the number of different arrangements.