How do you simplify #(9!)/(7!2!)#?

Question

7166 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-17T16:08:22

I got #36#

We can write: #9!"=9*8*7!#
So that our expression becomes:
#(9*8*cancel(7!))/(cancel(7!)2!)=(9*8)/(2*1)=9*4=36#

Answer 2 · 2016-08-17T16:08:29

#(9!)/(7!2!)=36#

#(9!)/(7!2!)=(9xx8xxcancel(7xx6xx5xx4xx3xx2xx1))/cancel(7xx6xx5xx4xx3xx2xx1)xx1/(2xx1)# #=# #72/2# #=# #36#