How do i solve this?

Question

#evaluate (3/3)# in factorial form

905 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-13T22:47:39

#((3),(3))=(3!)/(3!(3-3)!)#

Assuming you meant #((3),(3))#, this is another way to write #3C3#.

Also called a combination, #nCr=(n!)/(r!(n-r)!)#

Here, #n=3# and #r=3#.

Therefore, #((3),(3))=(3!)/(3!(3-3)!)#

Just in case you're curious about what this is equal to:

#=>(3!)/(3!(3-3)!)#

#=>(3!)/(3!(0)!)#

#=>(cancel(3!))/(cancel(3!)*1)#

#=>1#