How do you know that 8C2+8C3+8C4+8C5+8C6+.... = (2^8 - 8C0 - 8C1) ?

it is mentioned that at least 2 colors so we cannot ignore the possibilities of 8c3+8c4+...............

TAKE THREE COLRS RED BLUE GREEN ....(RGB)...RECALL SETS CHAPTER
NUMBER OF SUBSET WITH 3 ELEMENTS IS GIVEN BY 2^3=8 SUBSET...THIS INCLUDES A NULL SET WHERE NO COLOR IS CHOSEN...USE THIS ANALOGY
CHOOSING NO COLRS FROM SET(NC0=1) + CHOOSING ONE COLR (NC1=N)+(X=TWO COLOR+THREE COLOR.....)=2^N
NOW X IS GIVEN AS LESS THAN 500 SO JUST PUT THESE VALUES AND SOLVE
1+N+X=2^N
X=2^N-1-N
500<2^N-1-N
FROM OPTIONS ONLY SUITABLE VALUE IS 8..SO N=8

not clear with wordings as we can mix 2 colors or 3 colors and so on ...

in these types of Qs,
since we have to mix 2 or more colors,
let us assume if we have 3 colours R, G, B

so out of these 3 if we want to mix 2 or more, we can do like this

RG+GB+RB+RGB=4 Combination {Combination counts not permutation}
so selecting, 2 or more colors out of 3 is like
3c2+3c3=3+1

so same as if we have 9 colors
then we would go like this
9C2+9C3.....+9C9,
there is one famous formula for such series which is
nC0+nC1+nC2...........+nCn=n raised to the power 2
so it will give us 2 raised to the power 9 which is 512, {ofcourse we have to remove the combination 9c0 and 9c1} but still the mixtures will be more than 500 hence 8 must be the answer.