flaw in (2) may be. Not in the signs, but rather, in the lack of a term. A term that shouldn´t be a constant, but dependent of n . A term that added up to the other two would not only make N$_{3}$ possitive for n >> 1, but actually equal to six. So there must be an aditional source of degrees of freedom which Einstein missed to consider. If we could iden- tify where this source of degrees of freedom was, we would have our problem solved. Now, if we choose two particles, different from the ones used to generate the plane, the line that joins their centers of mass is a possi- ble rotation axis for S$_{3}$ . And since the num- ber of ways in which pairs may be chosen from a set of n − 3 particles is C 2 n − $_{3}$= ( n − 3)! 2!( n − 5)! = ( n − 3)( n − 4) 2 , (6) for n ≥ 3 . There will be an equal number of such axes. Each of this axes forms with the di- rection of the vector v an angle ϕ$_{i}$ which is a function of time and determines a possi- ble rotation of the system. In general, the different ϕ$_{i}$ will not hold relations of linear independence. A group of n particles may rotate in space without dissatisfying the condition that the distances between the particles remain con- stant. However, it is meanigless to talk about rotations without first establishing an ade- quate reference frame. To do so we arbi- trarily selected three particles from S$_{3}$ ; the points were the centers of mass of these par- ticles are located generate a plane P in three- dimensional space. And the vector v , which is orthogonal to P , designates an arbitrary direction in space. We must point out that we are defining v as a fixed vector, and that it is perpendicular to P in its original position, but as S$_{3}$ rotates, this perpendicularity rela- tion will be lost. Therefore, it is convenient to make a copy of P , which we will call P ´, and hold this copy fixed in the original position of P . Thus v will allways be orthogonal to P ´. By considering the plane P ´ and its normal vector, we are defining a three-dimensional coordinate system. However, (2) seems to be the correct ex- pression for n = 2. It also works for n = 3 and n = 4, which is not surprising, since for this value of n the last term in expression (7) is equal to zero, so (7) and (2) are equivalent. Once we had developed this expressions, we were curious on wether, by following the We believe that the number of ϕ$_{i}$ allowed to S$_{3}$ for a given value of n is the term miss- ing in Einstein’s calculation, and we propose that the number of degrees of freedom for the system S$_{3}$ is given by: N$_{3}$ = 3 n $_{−}$n ( n − 1) 2 + ( n − 3)( n − 4) 2 = 6 , (7) when n ≥ 3 . 4