Awesome, thank you very much.After reading through the wiki page, since n start from 0, I think the correctly answer for 100 numbers should be: 100!-------------- = 1,978,261,657,756,160,653,623,774,456 51! * 50! where n = 50.

I don't think that's correct.For i = 50, there are total of about 10^15 combination. Number of sequences such that the subtotal is always >= 0 should be much smaller. 50!------------ = 126,410,606,437,752 (about 10^15)25! * 25!

<t>Hello,I would like to get some help with following brainteaser:Suppose there are fifty "1" and fifty "-1".How many way this sequence of 100 numbers can be constructed such that the subtotal is always >= 0?For example:In case of 2 numbers, there is only 1 way:1, -1In case of 4 numbers, there are 2...