Remember that #n!# gives the product of all the numbers from 1 through to #n#.
#8! = 8xx7xx6xx5xx4xx3xx2xx1#
If you were to have #(8!)/(5!)# this is the same as:
#(8xx7xx6xxcancel(5!))/(cancel(5!))# which cancels to #8xx7xx6#
Simplifying manually is long and tedious, but here goes .....
#(15!)/(9!6!) + (7!)/(10!5!)" "larr# split up the bigger numbers.
Let's look at the first fraction... #(15!)/(9!6!)#
#(cancel15^cancel3xxcancel14^7xx13xxcancel(12)^cancel2xx11xxcancel10^5xxcancel(9!))/(cancel(9!)cancel(6)xxcancel5xxcancel4^cancel2xxcancel3xxcancel2xx1)" "larr# cancel
Cancel: #9! and 9!, " "6 " into "12 " twice, " "5 into " 15 # thrice,
#2 " into " 14 and 10 and 4," " 3 and 3#
This leaves #(7xx13xx11xx5)/1 = 5005#
Similarly for he second fraction:
#(7!)/(10!5!)#
# = (7xx6xxcancel(5!))/(10!xxcancel(5!))#
=#cancel(7xx6)/(10xx9xx8xxcancel(7xx6)xx5xx4xx3xx2xx1)#
=#1/(86400)#
Adding the two answers gives: #5005 + 1/86400#
#= 5000 1/86400#
