Delete any $d-1$ coordinates. The resulting puncturing map
$$
\pi:C\longrightarrow\mathbb F_2^{n-d+1}
$$
is injective: if $\pi(x)=\pi(y)$ for distinct codewords, then the nonzero codeword $x-y$ is supported on at most $d-1$ coordinates, contradicting the minimum distance. Thus
$$
2^k=|C|\leq2^{n-d+1},
$$
which is the <Singleton bound>
$$
\boxed{n\geq d+k-1}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 11G / a / 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>:
$$
\boxed{A(n,d)\le2^{n-d+1}}.
$$

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 1 / 11I / a / iii / Solution

{c}
{wiki}

For a code of length $n$, size $M$, and minimum distance $d$ over an alphabet of size $q$, puncturing $d-1$ coordinates gives the Singleton bound
$$
M\leq q^{n-d+1}.
$$


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 Incoming links: Singleton bound