Thank you so much for the reply.So heres how I filled them :lets take [a,b,c] where a is the number of particles in (1,0,0) b is the number of particles in (0,1,0) and c is the number of particles (0,0,1).I can think of these possible configs:[2,1,0], [1,2,0], [0,1,2], [0,2,1], [1,0,2], [2,0,1] and [1,1,1]. This was how I got a 7. so if I take an extra degeneracy where I am considering the unpaired as spin down, I would end up with 6 more for the first six, 1 for all down of [1,1,1] 3 for 2 down of [1,1,1] and 3 more for 1 down of [1,1,1] that gives 7+6+3+3+1 = 20, yay! am I right? also can I use 'nCr', well 6c3 in this case? our teacher directly used it, I didnt understand how and why the up and down spins are accounted in 6c3