Question

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?

Answer 1

`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 ocolor(white)(o)o o o ocolor(white)(o)o o o ocolor(white)(o)o o o`
`o o o ocolor(white)(o)o o o ocolor(white)(o)o o o ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)o`

`o o o ocolor(white)(o)o o o ocolor(white)(o)o o ocolor(white)(o)o o ocolor(white)(o)o`
`o o o ocolor(white)(o)o o o ocolor(white)(o)o o ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)o`
`o o o ocolor(white)(o)o o o ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)o`

`o o o ocolor(white)(o)o o ocolor(white)(o)o o ocolor(white)(o)o o ocolor(white)(o)ocolor(white)(o)o`
`o o o ocolor(white)(o)o o ocolor(white)(o)o o ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)o`
`o o o ocolor(white)(o)o o ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)o`
`o o o ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)o`

`o o ocolor(white)(o)o o ocolor(white)(o)o o ocolor(white)(o)o o ocolor(white)(o)o o o`
`o o ocolor(white)(o)o o ocolor(white)(o)o o ocolor(white)(o)o o ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)o`
`o o ocolor(white)(o)o o ocolor(white)(o)o o ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)o`
`o o ocolor(white)(o)o o ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)o`
`o o ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)o`
`ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)ocolor(white)(o)o`

I count `15` arrangements.

Answer 2

See below.

Explanation:

Calling `{alpha,beta,gamma} in ZZ ge 0` we have

`3alpha+2beta+gamma=15`

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