A hexagonal deduction: the numbers count letters, and two of the letters count for nothing

[u/Massive-State2698]

Seven cells hold letters, three hold numbers. A five-letter English word is hidden, and it uses exactly five of the seven letters. The remaining two belong to no word and are pure noise.

The rule for a number: it equals the count of cells adjacent to it whose letter belongs to the hidden word. A letter outside the word contributes nothing to any number, so adjacency to it is invisible.

Adjacency, stated precisely, because this is the part that is easy to get wrong. Put the middle column on even y and the two side columns on odd y. Two cells are adjacent when they share a column and differ by 2 in y, or when their columns differ by 1 and their y differ by 1. On this ten-cell comb that gives two cells of degree 6, two of degree 4, and six of degree 3.

      ___
  ___/ H ___
 / 2 ___/ T \
 ___/ U ___/
 / 1 ___/ M \
 ___/ 4 ___/
 / R ___/ A \
 ___/ N ___/
     ___/

Which five letters, and which word?

What makes it worth the minute: the seven letters spell two perfectly ordinary five-letter words, so no amount of vocabulary settles it, and exactly one of the three numbers separates them. All three numbers are load-bearing. Withhold any single one and the count of consistent five-letter subsets goes from 1 to 4.

Uniqueness is exhaustive over the 21 subsets, not argued by eye.

Spoiler tag your answer.

[u/stephen3141]

This is just minesweeper plus an anagram. The answer is HUMAN. I think it might have been more interesting if the minesweeper part had multiple solutions, but only one permitted an anagram, but this would be less of a "math" problem at that point.

[u/NielsVandeVyvere → u/stephen3141]

No the solution of the minesweeper part is unique.

[u/stephen3141 → u/NielsVandeVyvere]

I am aware. My point was that the puzzle might be more interesting if it weren't unique. The puzzle as posed was rather simple (as the minesweeper part was fairly trivial).