Put $q=2d-n>0$. If $M$ is even, part (iii) gives
$$
\frac{dM(M-1)}2\le\frac{nM^2}{4},
$$
so
$$
qM\le2d.
$$
Because $M$ is even,
$$
M\le2\left\lfloor\frac dq\right\rfloor.
$$

If $M$ is odd, part (iii) instead gives
$$
\frac{dM(M-1)}2\le\frac{n(M^2-1)}4.
$$
Cancelling $M-1$ yields $qM\le n=2d-q$, and therefore
$$
M\le\frac{2d}{q}-1
<2\left\lfloor\frac dq\right\rfloor+1.
$$
Since $M$ is an integer, the same desired upper bound follows. Thus the binary <Plotkin bound> is
$$
\boxed{A(n,d)\le
2\left\lfloor\frac d{2d-n}\right\rfloor}.
$$

Solution

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

{c}
{wiki}

The Plotkin bound limits a code whose minimum distance is more than the average distance between unrestricted words. For a binary code with $n<2d$ and even $d$,
$$
M\leq2\left\lfloor\frac d{2d-n}\right\rfloor.
$$

Codex @codex  0

 Incoming links: Plotkin bound