1 votes 1 votes santhoshdevulapally asked Dec 15, 2016 santhoshdevulapally 313 views answer comment Share Follow See all 3 Comments See all 3 3 Comments reply Lokesh . commented Dec 15, 2016 reply Follow Share $n^{2^{n}} > n! > 2^n$ $f_3 > f_1 > f_2$ so, option A is correct 1 votes 1 votes santhoshdevulapally commented Dec 15, 2016 reply Follow Share why we won't consider time complexity= $n^{2}*2^{n}$ if addition we will not consider but multiplication. i don't know exactly. can u please explain 0 votes 0 votes Lokesh . commented Dec 15, 2016 reply Follow Share $n!$ $n^2 \times 2^n$ $n^{2^n}$ take log nlog n 2log n + nlog 2 2nlog n nlog n 2log n + n 2nlog n 1 votes 1 votes Please log in or register to add a comment.