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

2 Answers
Apr 26, 2018

The result is #840#.

Explanation:

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:

https://www.desmos.com/calculator

Hope this helped!

Jul 8, 2018

#840#

Explanation:

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!