How do you simplify #(7!)/(3!)#?

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Answer 1 · 2018-04-26T02:04:37

The result is #840#.

You could use a calculator and actually expand it all out, or you could use the recursive definition of the factorial:

#x! =x*(x-1)!#

For instance, #8! =8*7!#. In our problem, you will see that eventually, things cancel out.

#color(white)=(7!)/(3!)#

#=(7*6!)/(3!)#

#=(7*6*5!)/(3!)#

#qquadqquadqquadvdots#

#=(7*6*5*4*3!)/(3!)#

#=(7*6*5*4*color(red)cancelcolor(black)(3!))/color(red)cancelcolor(black)(3!)#

#=7*6*5*4#

#=42*5*4#

#=210*4#

#=840#

You can use a calculator to check your answer:

Hope this helped!

Answer 2 · 2018-07-08T23:42:24

#840#

Recall that #7! =7*6*5*4*3*2*1# and

#3! =3*2*1#

This allows us to rewrite #(7!)/(3!)# as

#(7*6*5*4*3*2*1)/(3*2*1)#

We cancel out common factors on the top and bottom to get

#(7*6*5*4*cancel(3*2*1))/cancel(3*2*1)#

#=>7*6*5*4=840#

Hope this helps!