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?

2 Answers
Feb 12, 2017

Answer:

#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.

Feb 12, 2017

Answer:

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.