Solution
= Solution
Iterate the <punctured code>[puncturing] inequality from part (ii) $d-1$ times and then use part (i):
$$
A(n,d)
\le A(n-1,d-1)
\le\cdots\le A(n-d+1,1)
=2^{n-d+1}.
$$
This is the binary <Singleton bound>:
$$
\boxed{A(n,d)\le2^{n-d+1}}.
$$