Not the number of $n$ 's; the number of new lines

$Let\quad AB\quad be\quad a\quad line\quad and\quad let\quad it\\ be\quad intersected\quad by\quad CD\quad at\quad a\quad point\quad P.\\ Clearly,\quad P\quad is\quad one\quad of\quad the\quad n-1\quad points\\ formed\quad by\quad AB's\quad intersection\quad with\quad the\\ other\quad lines.\quad So\quad the\quad aggregate\quad of\quad all\quad \\ such\quad points\quad =\quad n(n-1).\quad But\quad each\quad such\\ point\quad would\quad be\quad counted\quad twice.\quad So,\\ the\quad total\quad number\quad of\quad points=\frac { n(n-1) }{ 2 } .\\ \\The\quad number\quad of\quad points\quad of\quad intersection\\ on\quad AB\quad and\quad CD\quad =\quad 2(n-1)-1.\quad If\quad P\quad is\\ joined\quad to\quad any\quad of\quad these\quad points,\quad then\\ it\quad would\quad not\quad count\quad as\quad a\quad new\quad line.\\ \therefore \quad Number\quad of\quad points\quad that\quad P\quad can\quad join\\ so\quad that\quad the\quad line\quad so\quad formed\quad counts\quad as\quad \\ a\quad new\quad line\quad =\quad \frac { n(n-1) }{ 2 } -(2n-3) \quad = \quad \frac { (n-2)(n-3) }{ 2 } .\quad \\ \\ So\quad the\quad aggregate\quad of\quad all\quad such\quad lines\quad =\quad \frac { n(n-1) }{ 2 } (\frac { (n-2)(n-3) }{ 2 } )\\ But\quad once\quad again,\quad all\quad the\quad lines\quad are\quad counted\quad twice.\\ \Rightarrow \quad Total\quad number\quad of\quad new\quad lines\quad =\quad \frac { 1 }{ 8 } n(n-1)(n-2)(n-3).\\ \therefore \quad |a|+|b|+|c|+|d|+|e|+|f|=\boxed { 15. }$