Show $A_q(n,n-1)=q$, if $n > \binom{q+1}{2}$

Question

See title: Show $A_q(n,n-1)=q$, if $n>\binom{q+1}{2}$, where $A_q(n,n-1)$ is the maximum number of words of length $n$ a code can have, over a $q$-ary alphabet, with minimum Hamming distance at least $n-1$. I try to count, but my counting is very rusty and I cannot get the desired result. Thanks!

Answer 1

Let $C$ be a set of $q+1$ words, and suppose by way of contradiction that each pair of words agree in at most one position. There are $\binom{q+1}2$ pairs of words, and $n$ positions. Since $n>\binom{q+1}2$, there would exist a position such that no two codes agree in that position. Can you take it from here?