For strictly increasing f, if we choose elements a1<a2<⋯<a6, then f(i)=ai.
Since f(i)≥i always, condition f(i)=i means f(i)≥i+1 for all i.
Let h(i)=f(i)−i. Then h(i)≥1, h(6)≤3, and h is non-decreasing.
So h(i)∈{1,2,3} for all i.
Number of non-decreasing sequences of length 6 from {1,2,3}=(28)=28.