A number theory problem by Zarak K

$So\quad first\quad of,\quad we\quad can\quad simply\quad take\quad two\quad numbers\quad at\quad a\quad time\quad \\ and\quad it'll\quad all\quad give\quad different\quad solutions.\\ \\ Now\quad here's\quad the\quad tricky\quad part.\quad From\quad the\quad second\quad step,\quad \\ we\quad disclude\quad 0\quad from\quad the\quad set\quad since\quad it\quad will\quad give\quad \\ the\quad same\quad result\quad as\quad the\quad previous\quad step.\\ \\ So\quad now\quad we\quad take\quad 3\quad numbers\quad at\quad a\quad time,\quad then\quad 4\quad at\quad time\\ \quad and\quad finally\quad 5\quad at\quad a\quad time.\\ \\ Therefore\quad we\quad have:-\quad \\ \\ \quad \quad \quad \quad \quad \quad \quad _{ 2 }^{ 6 }{ C }\quad +\quad _{ 3 }^{ 5 }{ C }\quad +\quad _{ 4 }^{ 5 }{ C }\quad +\quad _{ 5 }^{ 5 }{ C }\\ =\quad \quad 15\quad +\quad 10\quad +\quad 5\quad +\quad 1\quad \\ =\quad \boxed { 31 } \\ \\ NOTE:-\quad We\quad observe\quad that\quad solution\quad in\quad all\quad r\quad \\ is\quad different\quad because\quad sum\quad of\quad any\quad \\ numbers\quad is\quad not\quad equal\quad to\quad another\quad number.\quad \quad \quad \quad \quad \\$

First see that all the numbers in the set are powers of 2 . we know that every number can be represented as sum of powers of 2 . Therefore the smallest and the largest number that can be formed are 1 and 1+2+4+8+16 that is 1 and 31 . Therefore number of solutions is 31 .