Three numbers a, b and c are chosen at random (simultaneously) from among the numbers 1, 2, 3, ..., 99. The probability that a3 + b3 + c3 - 3abc is divisible by 3, is

\(\rm\frac{3\times {^{33}C_3}- \left(^{33}C_1\right)^3}{^{99}C_3}\)

\(\rm\frac{3\times {^{33}C_3}+ \left(^{33}C_1\right)^3}{^{99}C_3}\)

\(\rm\frac{2\times {^{33}C_3}+ \left(^{33}C_1\right)^3}{^{99}C_3}\)

\(\rm \frac{2\times {^{33}C_3}-\left(^{33}C_1\right)^3}{^{99}C_3}\)

Three numbers a, b and c are chosen at random (simultaneously) from am

1.	Three numbers a, b and c are chosen at random (simultaneously) from among the numbers 1, 2, 3, ..., 99. The probability that a3 + b3 + c3 - 3abc is divisible by 3, is
A.	\(\rm\frac{3\times {^{33}C_3}+ \left(^{33}C_1\right)^3}{^{99}C_3}\)
B.	\(\rm\frac{3\times {^{33}C_3}- \left(^{33}C_1\right)^3}{^{99}C_3}\)
C.	\(\rm\frac{2\times {^{33}C_3}+ \left(^{33}C_1\right)^3}{^{99}C_3}\)
D.	\(\rm \frac{2\times {^{33}C_3}-\left(^{33}C_1\right)^3}{^{99}C_3}\)
Answer» B. \(\rm\frac{3\times {^{33}C_3}- \left(^{33}C_1\right)^3}{^{99}C_3}\)