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
