How do you simplify #(16!)/(9!*7!)#?

Question

3869 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-31T14:57:11

#16! =1*2*3*4*5*6*7.......*16#

#9! =1*2*3*4*5*6*7...*9#

#7! =1*2*3*4*5*6*7#

#(16!)/(9!*7!)=#

#(1*2*3*4*5*6*7*8*9*10*11*12*13*14*15*16)/((1*2*3*4*5*6*7*8*9)(1*2*3*4*5*6*7))#

The number that equal other cancel out!

#rarr=(10*11*12*13*14*15*16)/(1*2*3*4*5*6*7)#

#7# and #14# cancel out ,#6# and #12# cancel out ,#5# and #10# cancel out!

#rarr=(2*11*2*13*2*15*16)/(1*2*3*4*1*1*1)#

Now #4# and #14# cancel out,#2# and #2# cancel out, #3# and #15# cancel out!

#rarr=(2*11*1*13*2*5*4)/(1*1*1*1*1*1*1)#

#rarr=2*11*1*13*2*5*4#

#rarr=22*26*20#

#rarr=440*26#

#rarr=11440#

Answer 2 · 2016-02-07T21:21:53

11440

#(16!)/(9!*7!)=( 16*15*14*13*12*11*10*cancel(9!))/(cancel(9!)*7!)#

#( 16*cancel(15)*14*13*12*11*10)/(7*6*cancel(5)*4*cancel(3)*2*1)#

#( 16*cancel(14)*13*12*11*10)/(cancel(7)*6*4*cancel(2)*1)#

#(16*13*6*2*11*(2*5))/(6*2*2)#

#16*13*11*5#

11440