I'm studying the combinatorics "twelvefold way", and found an identity that cannot explain myself.
The case, $$ {x-1 \choose b-1} $$ as far as I understand is derived the following way: $$ {(x-b)+b-1 \choose x-b}={x-1 \choose x-b}={x-1 \choose b-1} $$ The first identity is evident, but I don't understand the second one: $$ {x-1 \choose x-b}={x-1 \choose b-1} $$ Why x-b is equal to b-1?
Thanks