Prove that:

n! / r! (n-r)! + n! / (r-1)! (n-r+1)! = (n+1)1 / r!(n-r+1)!

see first keywords to solve,

  1.  r!=(r)(r-1)!
  2. (n-r+1)!=(n-r+1)(n-r)!

now,

to prove,

n! / r!(n-r)! + n! / (r-1)! (n-r+1)! = (n+1)! / r!(n-r+1)!

L.H.S,

n! / (r)(r-1)! (n-r)! +n! / (r-1)! (n-r+1)(n-r)!

n! / (r-1)! (n-r)! will be common

so,

n! / (r-1)! (n-r)! *[ 1/r  +  1/(n-r+1) ]

n! / (r-1)! (n-r)! *[ n-r+1+r / nr-r² +r ]

we will get ,

n! / (r-1)! (n-r)! / * [n+1 / r(n-r+1)]

(n!) (n+1) / (r-1)! (r) (n-r)! (n-r+1)

(n+1)! / r! ( n-r+1)!

L.H.S=R.H.S