# Totally antisymmetric part of a rank 3 tensor

In Hobson's General Relativity: An Introduction for Physicists book, pg. 95, the totally antisymmetric part of a rank 3 tensor $$t_{abc}$$ is defined as

$$t_{[abc]} = \frac{1}{6}(t_{abc}-t_{acb}+t_{cab}-t_{cba}+t_{bca}-t_{bac}).$$

Why did the author include a factor of $$1\over 6$$ in the expression? Is he trying to anticipate that the tensor $$t_{abc}$$ can be made up in some way using the totally antisymmetric tensor $$t_{[abc]}$$?

Adding the $$\frac{1}{n!}$$ factor in the antisymmetrisation is just a convention (and some authors omit it) - but this convention ensures that for totally antisymmetric tensors, $$T_{\mu\nu...\sigma} = T_{[\mu\nu...\sigma]}$$.