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                     Now, similarly

                                                           (n − k)!


                     is just (n − k − 1)! with one more layer


                                             (n − k)! =(n − k)(n − k − 1)!.


                     So we can put the multiplicative identity 1 in disguise to re-write


                                                n!                n!
                                                       +
                                            k!(n − k)!   (k +1)!(n − k − 1)!

                                           n!     k +1             n!          n − k
                                    =                   +
                                       k!(n − k)! k +1    (k +1)!(n − k − 1)! n − k


                                                (k +1)n!           (n − k)n!
                                          =                  +
                                            (k +1)!(n − k)!     (k +1)!(n − k)!


                        which now can be written in one fraction as


                  (k +1)n!+ (n − k)n!      kn!+ n!+ nn! − kn!          (n +1)n!            (n +1)!
                                        =                        =                  =                  .
                    (k +1)!(n − k)!          (k +1)!(n − k)!       (k +1)!(n − k)!     (k +1)!(n − k)!


                        This means that


                                 n!                          n!                         (n +1)!
                                                                               b
                    a k+1 n−k           + a k+1 n−k                     = a k+1 n−k                 .
                         b
                                               b
                             k!(n − k)!             (k +1)!(n − k − 1)!             (k +1)!(n − k)!






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