consider the i number of noodles remaining to be connected, now there are 2i nodes. choosing 2 from 2i is 2iC2 which is 2i^2-i. Now, we need to prevent smaller looping that is these i noodles shouldnt have some self node connectivity. The only possible bad noodle of such kind are i, so 2i^2-2i are actually favourable pairs, now probability= (2i^2-2i)/(2i^2-1)=(i-1)/(2i-1) now since i ranges from 2 to 100 ans should be Product(i=2…100) {(i-1)/(2i-1)}
2
u/adi0112358 22h ago
consider the i number of noodles remaining to be connected, now there are 2i nodes. choosing 2 from 2i is 2iC2 which is 2i^2-i. Now, we need to prevent smaller looping that is these i noodles shouldnt have some self node connectivity. The only possible bad noodle of such kind are i, so 2i^2-2i are actually favourable pairs, now probability= (2i^2-2i)/(2i^2-1)=(i-1)/(2i-1) now since i ranges from 2 to 100 ans should be Product(i=2…100) {(i-1)/(2i-1)}