In this paper we prove the following:

1. 1. Let $m\ge 2,\,n\ge 1$ be integers and let $G$ be a group such that ${{(XY)}^{n}}\,=\,{{(YX)}^{n}}$ for all subsets $X,Y$ of size $m$ in $G$ . Then

   1. a) $G$ is abelian or a $\text{BFC}$ -group of finite exponent bounded by a function of $m$ and $n$ .

   2. b) If $m\ge n$ then $G$ is abelian or $|G|$ is bounded by a function of $m$ and $n$ .

2. 2. The only non-abelian group $G$ such that ${{(XY)}^{2}}\,=\,{{(YX)}^{2}}$ for all subsets $X,Y$ of size 2 in $G$ is the quaternion group of order 8.

3. 3. Let $m$ , $n$ be positive integers and $G$ a group such that

   $${{X}_{1}}\cdot \cdot \cdot \,{{X}_{n}}\,\subseteq \,\bigcup\limits_{\sigma \in {{S}_{n}}\,\backslash \,1}{{{X}_{\sigma (1)}}\cdot \cdot \cdot \,{{X}_{\sigma (n)}}}$$

for all subsets ${{X}_{i}}$ of size $m$ in $G$ . Then $G$ is $n$ -permutable or $|G|$ is bounded by a function of $m$ and $n$ .