30 votes 30 votes The number of binary strings of $n$ zeros and $k$ ones in which no two ones are adjacent is $^{n-1}C_k$ $^nC_k$ $^nC_{k+1}$ None of the above Combinatory gate1999 combinatory normal + – Kathleen asked Sep 23, 2014 • edited May 16, 2018 by kenzou Kathleen 9.0k views answer comment Share Follow See all 8 Comments See all 8 8 Comments reply Show 5 previous comments Venkat Sai commented Dec 25, 2017 i edited by Venkat Sai Dec 25, 2017 reply Follow Share but that's the actual gate question mentioned with the condition so I asked him to edit and if the mention that condition the possibility we have to consider all the possibilities for n=k-1,k,k+1..... and that's what i wrote above anyways for both we get none of the above ...they didn't mention n as a fixed value in the question if u don't care about what they gave in the book u can see from the original gate paper of 1999.....but i dont think its available anywhere but if they donot mention that condition we dont need to do OR for all the possible cases.. i 0 votes 0 votes sk26 commented Oct 20, 2020 reply Follow Share Please can anyone explain why only case of n distinct zeroes being considered? we could also have a case like, 10010001. In that case C(n+1, k) wont be true. Please comment 0 votes 0 votes kiioo commented Nov 10, 2020 reply Follow Share Distinct doesn't mean that the zeroes can’t be adjacent to each other. We just need that no two ones are together. 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes Since there are n zeroes, so XOXOXOXOXOXOX n+1 gaps can be possible, where 1's can be placed so that no two one's are adjacent. So, no. of ways in which k 1's can be placed in n+1 gaps are, $^{n+1}$$C_k$ Hence Option “D” None of these is the answer varunrajarathnam answered Oct 14, 2020 • edited Apr 23, 2021 by varunrajarathnam varunrajarathnam comment Share Follow See all 0 reply Please log in or register to add a comment.