The Tetrahedron has 4 triangular surfaces with 4 vertices/corners ( say A,B,C and D) as can be seen here. http://www.sjsu.edu/faculty/wa... Now if you take a point inside a tetrahedron (suppose O) and connect it with any two of its corners which are nothing but vertices( suppose A and B), you will get 1 internal plane as OAB. So we can see from here that, no of new internal planes = no of different pair of corners or vertices Similarly you can take any other 2 corners like (A,C) or (A,D) or (B,C) or (B,D) or (C,D), hence total possible pair of corners are 6. Therefore 6 new internal planes possible. We could also calculate the possible corners by using combinations formula, which is nCr, i.e. no of ways to select a combination of r things from a given set of n things. here n = 4 ( as total 4 vertices, A,B,C and D) and r =2 ( as we need two corners at a time) Thus, 4C2 = 6.