How do you simplify #(15!)/(2!7!)#?

1 Answer
Steve M
Dec 13, 2016

We can rewrite as
# 15! = 15xx14! = 15xx14xx13! .. # and so on
So #15=15xx14xx13xx12xx11xx10xx9xx8xx7!#

And so,
# (15!)/(2!7!) = (15xx14xx13xx12xx11xx10xx9xx8xx7!)/(2!7!) #
# \ \ \ \ \ \ \ \ = (15xx14xx13xx12xx11xx10xx9xx8)/(2!)#
# \ \ \ \ \ \ \ \ = (15xx14xx13xx12xx11xx10xx9xx8)/(2)#
# \ \ \ \ \ \ \ \ = 129739600#

Related questions
See all questions in Factorial Identities
Impact of this question
1481 views around the world
You can reuse this answer
Creative Commons License