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Question 1
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Three numbers a,b and c are chosen at random (without replacement) from among the numbers 1, 2, 3, ..., 99. The probability that $a^3+b^2+c^2-3abc$ is divisible by 3 is,
A.
$$ \frac{3.^{33}C_3+(^{33}C_1)^3}{^{99}C_3} $$
B.
$$ \frac{3.^{33}C_3-(^{33}C_1)^3}{^{99}C_3} $$
C.
$$ \frac{2.^{33}C_3+(^{33}C_1)^3}{^{99}C_3} $$
D.
$$ \frac{2.^{33}C_3-(^{33}C_1)^3}{^{99}C_3} $$
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