= Solution
The <Pauli exclusion principle> requires the complete <electron> <wavefunction> to change sign on exchange of any two <electrons>, including both position and <spin> labels. Let $a=1s$ and $b=2s$ be normalized, orthogonal spatial <atomic orbitals>, and let $\alpha,\beta$ denote <spin> up and down.
The symmetric and antisymmetric spatial functions are
$$
\Phi_\pm=\frac{a(1)b(2)\pm b(1)a(2)}{\sqrt2}.
$$
The <singlet state> is
$$
\boxed{\Psi_{S=0}=\Phi_+\,
\frac{\alpha(1)\beta(2)-\beta(1)\alpha(2)}{\sqrt2}.}
$$
The three <triplet states> share the antisymmetric spatial factor:
$$
\boxed{\begin{aligned}
\Psi_{1,+1}&=\Phi_-\alpha(1)\alpha(2),\\
\Psi_{1,0}&=\Phi_-\frac{\alpha(1)\beta(2)+\beta(1)\alpha(2)}{\sqrt2},\\
\Psi_{1,-1}&=\Phi_-\beta(1)\beta(2).
\end{aligned}}
$$
Every factor and complete product is normalized. Here the subscripts are $(S,M_S)$; the spatial <electron configuration> is $1s2s$ in each case.
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