let the no of 1's be = a no of 0's = k total permutations are (a+k)!/(a!*k!) - * now consider the case where at least two 0's are conseq. then, total elements are (assuming 00 to be 1 element) =(a+k-1) total permutations are (a+k-1)!/(a!* [k-1]!) - ** subtracting ** from * gives our answer