How do you simplify #(21!)/(17!4!)#?
2 Answers
Explanation:
#(21!)/(17! xx 4!)=(17! xx 18 xx 19 xx 20 xx 21)/(17! xx4!)#
may be simplified as
#=(18 xx 19 xx 20 xx 21)/(1 xx 2 xx 3 xx 4)#
and this may be further simplified as
#=(9 xx 19 xx 5 xx 7)/(1 xx 1 xx 1 xx 1)#
#=9 xx19 xx 5 xx 7#
#=5985#
Write out the definitions of each factorial and you should get
#=> (1cdot2cdot3cdot4cdots17cdot18cdot19cdot20cdot21)/((1cdot2cdot3cdot4cdots15cdot16cdot17)(1cdot2cdot3cdot4))#
The portion of
#= (18cdot19cdot20cdot21)/(1cdot2cdot3cdot4)#
#= ((19-1)cdot19cdot20(20+1))/(24)#
#= ((361 - 19)cdot(400 + 20))/(24)#
#= (342cdot420)/(24)#
#= (171cdot420)/(12)#
#= (171cdot210)/(6)#
#= (57cdot210)/(2)#
#= 57cdot105#
#= 57cdot(100 + 5)#
#= 5700 + 57cdot10/2#
#= 5700 + 285#
#= color(blue)(5985)#
