Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Discrete mathematics, continuous mathematics, and probability / 8 / a / iv

The $m$ red positions form a uniformly chosen subset of size $m$, so there are $\left(\genfrac{}{}{0ex}{}{m+n}{m}\right)$ equally likely colour patterns. A contiguous red block can start at positions $1,2,\dots ,n+1$, giving exactly $n+1$ favourable patterns. ThereforeIf the cards are regarded as individually labelled, each colour pattern corresponds to $m!n!$ permutations, so the same counting argument still applies. In the edge case $m=1$, the formula correctly gives probability one.