= Solution
For two equivalent $p$ <electrons>, antisymmetry permits
$$
\boxed{{}^3P_{0,1,2},\qquad{}^1D_2,\qquad{}^1S_0.}
$$
Their $9+5+1=15$ states exhaust $\binom62$. <Hund's rules> first maximize <spin>, then orbital <angular momentum>; the less-than-half-filled $p$ shell puts the smallest $J$ lowest. Hence the ground level is \b[${}^3P_0$]. This is the <allowed LS terms of a p-squared configuration>.<Energy Levels of Neutral Carbon by National Institute of Standards and Technology>[Ground configuration2s²2p² and its³P,¹D,¹S levels]{ref}
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