= Solution
The simplices in a <barycentric subdivision> are strict chains of nonempty faces. A tetrahedron has $15$ nonempty faces, so $f_0=15$. Counting strict two-face chains gives
$$
f_1
=\sum_{j=1}^4\binom4j(2^j-2)
=50.
$$
A strict three-face chain amounts to choosing a terminal face and partitioning its vertices into three nonempty ordered blocks. Hence
$$
f_2
=\binom43\,3!\,S(3,3)+\binom44\,3!\,S(4,3)
=24+36=60,
$$
where $S(-,-)$ denotes a Stirling number of the second kind. Therefore
$$
\boxed{\chi(|M|)=f_0-f_1+f_2=15-50+60=25}.
$$
Equivalently, the full subdivision has $4!=24$ tetrahedra and Euler characteristic one, so deleting its three-dimensional simplices from the alternating count gives $1+24=25$. This is the f-vector computation for the <barycentric subdivision of a tetrahedron>.
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