Count signals by their remaining colour multiplicities

Flags of the same colour are indistinguishable. The number of six-flag strings is $6!/(3!2!)=60$.

Six flags contribute 60. For five flags: omit green, giving $5!/(3!2!)=10$; omit one red, giving $5!/(2!2!)=30$; or omit one blue, giving $5!/3!=20$. Thus $60+10+30+20=120$ signals use at least five flags.

The source labels this approach “Not recommended”. Delete the final flag from each six-flag string. This gives 60 distinct five-flag strings: the colour missing from a five-flag string is uniquely determined by subtracting its multiplicities from the inventory (1 green, 3 red, 2 blue), so exactly one last colour restores it. Conversely every feasible five-flag string extends this way. Hence there are $60+60=120$ signals. This explains the source uniqueness claim rather than assuming deletion always preserves distinctness.