Why don't we have particles whose wavefunctions are symmetric wrt one exchange operator and anti-symmetric wrt other exchange operator? Consider a system with $n$ identical particles. Let the wavefunction of the system be $\psi(r_1,\ldots, r_2)$. Let $P_{a,b}$ represent the exchange operator which exchanges particle $a$ with particle $b$. Similarly, $P_{c,d}$  represent the exchange operator which exchanges particle $c$ with particle $d$. Now, suppose $P_{a,b}(\psi(r_1,\ldots, r_2))= - \psi(r_1,\ldots, r_2)$ and $P_{c,d}(\psi(r_1,\ldots, r_2))= \psi(r_1,\ldots, r_2)$. Why don't we have particles with wavefunctions satisfying the above property?
 A: It is not possible to have a state with four indistinguishable particles such that $P_{12} \psi = -\psi$ and $P_{34} \psi =\psi$, for an algebraic reason. Namely, the exchange operators have to form a representation of the permutation group $S_4$. It is rather well known that there are exactly two representations of $S_n$: the trivial representation where all exchange operators are $1$, and the parity representation where all single transpositions are represented by $-1$ and this is extended by the group law.
Thus either all $P_{nm} = 1$ or all $P_{nm} = -1$. Anything else is simply not consistent with the algebra of permutations.
A: A wavefunction is a normalised vector (or a ray) in the Hilbert space of vector states (I will assume finite degrees of freedom, so the C*-algebra of the system can be taken to be $B(H)$, with $H$ an $L^2(\mathbb R^n)$ space). Spin gives a superselection rule, and therefore there must be superselection sectors. It follows from the general theory that vector states from different sectors cannot be combined to yield another vector state, but just a statistical mixture. So a wavefunction is either bosonic or fermionic. To describe a state with mixed particles you need more general linear functionals which combine as convex combination of states from the superselection sectors.
