same line of reasoning, we could calculate the number of degrees of freedom of S$_{4}$ , that is, of the system made up by n particles with fixed relative distances, but which is, unlike S$_{3}$ , immerse in a four dimentional space. In this four-dimentional case, four coordi- nates are needed to locate the center of mass of each particle, which makes 4 n coordinates for the set of n particles. And the number of constraint equations is the same as for S$_{3}$ In principle, the number of degrees of freedom should be the same as for a tetra- dimensional rigid body. And in four dimen- sions there are ten degrees of freedom for the rigid body: four coordinates are needed to locate its center of mass and there are six possible rotation angles. Now, in the case of the n particles with fixed distances, we need 4 n coordinates to locate the particles´ cen- ters of mass, while the number of distances is still n ( n − 1) 2 . And the number of possible rota- tion angles is obtain observing that a ”hiper- plane” can be defined with four points and that the number of diferent ways in which pairs may be chosen from a group of n − 4 particles is given by: S$_{4}$ is and since the number of possible ϕ$_{i}$ is equal to zero for these values of n. That N$_{4}$ is equal to ten for any value of n less than or equal to four is consistent with the fact that ten is also the number of degrees of freedom of a rigid body in four-dimentional space (four coordinates are needed to locate the center of mass, and six more to describe the orientation of the body. Indeed, our pro- cedure works for the four-dimentional as it does for the three-dimensional case. More- over, we believe that it works for the general case. We propose that for a system of n par- ticles with fixed relative distances, immerse in a space of D dimensions, the number of degrees of freedom is given by: and by: Then, the number of degrees of freedom of 5 N$_{4}$ = 4 n $_{−}$n ( n − 1) 2 + ( n − 4)( n − 5) 2 = 10 , (9) when n ≥ 4 , N$_{4}$ = 4 n − n ( n − 1) 2 , (10) when 2 ≤ n ≤ 5 , N$_{D}$ = Dn $_{−}$n ( n − 1) 2 + ( n − D )( n − D − 1) 2 (11) = D ( D + 1) 2 , when n ≥ D, N$_{D}$ = Dn − n ( n − 1) 2 , (12) when 2 ≤ n ≤ D + 1 . C 2 n − $_{4}$= ( n − 4)! 2!( n − 6)! = ( n − 4)( n − 5) 2 , (8) for n ≥ 4 .