NYT Pips Hints

Every domino in the five-piece set carries a one, and one domino is a double one, so the set holds six ones in total. The board has a single region asking for six matching cells, which uses every one of them at once. That leaves a six, a five, a two and a zero for the four lone cells outside that region. One wants anything over five, which can only be the six; another wants anything over three, which is left to the five; and the smallest cell, wanting less than two, takes the zero, leaving the two for the last open space.

The eight dominoes hold four sixes, three fours, and two fives, and every one of those counts turns out to matter. Two lone cells demand anything over five, which only a six can satisfy, and a third six sits in a sum-to-seven pair alongside a one. The region asking for three identical cells uses all three fours in the set. The two sum-to-five pairs are built from the set's only two fives, each paired with a zero, and the less-than-three region absorbs most of what is left.

The fifteen-domino set has an unusually tidy structure: for several faces, the regions that need that number use up exactly the dominoes that carry it, no more and no less. Eight zeros in the set match a five-cell region that must all be zero, plus two smaller regions summing to nothing. Five sixes match a pair of high-sum regions, one totalling 16 and one totalling 17, plus two more regions elsewhere on the board. Six threes match three lone cells wanting exactly three, a matching pair, and a region summing to nine. Place those three families of pieces first, and the handful of remaining regions, mostly single cells with small targets, resolve without any guessing.