We have the general expression #a_n=2^n/(n!)# and are looking for #a_10#. That gives us:
#a_10=2^10/(10!)#
And now let's evaluate it! (I'll express the numbers 8 and 4 in terms of #2^x# to make cancellations and simplification more obvious):
#=(2^3xx2^2xx2xx2xx2xx2^2)/(10xx9xx2^3xx7xx6xx5xx2^2xx3xx2)#
#=(cancelcolor(red)(2^3)xxcancelcolor(brown)(2^2)xxcancelcolor(orange)2xxcancelcolor(green)2xxcancelcolor(blue)2xx2^2)/(cancelcolor(green)10^5xx9xxcancelcolor(red)(2^3)xx7xxcancelcolor(blue)6^3xx5xxcancelcolor(brown)(2^2)xx3xxcancelcolor(orange)2)#
#=4/(5xx9xx7xx3xx5xx3)=4/(14,175)#
