Exercise 3: A Different Split of the Same Scope — Possible Solution
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RESULTS
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    Original split [3, 3, 3, 2, 2] (sum=13): 9 points shipped
    New split [5, 4, 4] (sum=13): 9 points shipped

Both splits happen to ship the same 9 points here - but the split
itself, not just the point total, genuinely determines the outcome, as
a further check reveals:

    [5, 4, 4] -> 9 points shipped
    [4, 4, 5] -> 8 points shipped   (same three numbers, different order!)
    [4, 5, 4] -> 9 points shipped
    [6, 4, 3] -> 10 points shipped
    [7, 3, 3] -> 10 points shipped

Reordering the identical multiset {4, 4, 5} from [5,4,4] to [4,4,5]
drops the shipped total from 9 to 8 - the 5-point story never gets
started at all in that order, because 4+4=8 leaves only 2 points of
capacity remaining, not enough for the 5-point story, even though it
would have fit perfectly as the FIRST or SECOND item taken.

WHY THIS WORKS AS AN ANSWER
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This reveals something the chapter's own single example didn't fully
expose: "Small" isn't only about individual story size - the ORDER
stories are pulled in interacts with size to determine how much
capacity gets wasted at the boundary. A team that always pulls its
largest remaining story first (a real, common backlog-ordering
heuristic) would have avoided this exact trap - [5,4,4] and [7,3,3]
and [6,4,3] all ship more than [4,4,5] specifically because a
larger-first ordering leaves smaller, more flexible remainders to fill
in the leftover capacity, rather than leaving one large story stranded
at the end with too little room left for it.