2 votes 2 votes How many bit strings contain exactly eight 0s and 10 1s if every 0 must be immediately followed by a 1? I got answer 9C2=36.Answer given 45 Combinatory combinatory counting + – Anu asked Jul 15, 2015 • retagged Jun 27, 2017 by Arjun Anu 3.8k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 11 votes 11 votes Consider the following arrangement of eight 0 ans 1. _01_01_01_01_01_01_01_01_ Now the additional two 1's can be placed separately in 9C2 and placed together in 9 ways. So the answer is 9C2+9=36+9=45. Mari Ganesh Kumar answered Jul 15, 2015 • edited Jul 15, 2015 by Mari Ganesh Kumar Mari Ganesh Kumar comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments Gate Mm commented Dec 9, 2015 reply Follow Share Plz xplain this part Now the additional two 1's can be placed separately in 9C2 and placed together in 9 ways. 0 votes 0 votes Kaluti commented Sep 2, 2017 reply Follow Share why we are adding 9 to 36 here? 0 votes 0 votes Sanjay Sharma commented Jan 28, 2018 reply Follow Share these 9 cases includes 110101010101010101 011101010101010101 010111010101010101 010101110101010101 010101011101010101 010101010111010101. 010101010101110101. 010101010101011101 .010101010101010111 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes the bit strings must consist of eight 01 substrings and two 1s. Thus, there are ten total positions and choosing the two positions for the 1s determines the string. There are 10C2=45 such strings Rohan Ghosh answered Jul 15, 2015 Rohan Ghosh comment Share Follow See 1 comment See all 1 1 comment reply radha gogia commented Jan 27, 2016 reply Follow Share There are 9 positions for the remaining 2 1's ,how come 10 positions ? 0 votes 0 votes Please log in or register to add a comment.
