How do you write #60# using factorials?

1 Answer
Feb 3, 2016

#(5!)/(2!)#

Explanation:

First, try to find a series of consecutive integers whose product is #60#.

#60=2^2xx3xx5=5xx4xx3#

Note that #5xx4xx3# is the start of the #5!# factorial chain, without #2! =2xx1#.

#60=5xx4xx3=(5xx4xx3xx2xx1)/(2xx1)=(5!)/(2!)#

This is the simplest way to express #60#, but we can try in more exotic ways:

#60=10xx6=(10!)/(9!)xx(6!)/(5!)=((10!)(6!))/((9!)(5!))#