WebAnswer : 26since first and last bit have been already determined. How many 8-bit strings have either the second or the fourth bit 1 (or both)? Answer: 27+ 27- 26( # of 8-bit strings with second bit 1 plus the # of 8-bit strings with fourth bit 1 minus the # of 8-bit strings with both second and fourth bit 1 ) or WebDec 20, 2024 · We can think of each row as a 6-bit string of weight 3 (since of the 6 coins, we require 3 to be pennies). Thus there are \({6 \choose 3} = 20\) rows possible. Each row …

1.2: Binomial Coefficients - Mathematics LibreTexts

http://math.utep.edu/faculty/cmmundy/Math%202400/Exams/Some%20practice%20exam%202%20solutions.pdf Webkindly answer it perfectly Transcribed Image Text: Recall that a 5-bit string is a bit strings of length 5, and a bit string of weight 3, say, is one with exactly three 1's. How many 5-bit strings are there? How many 5-bit strings have weight 0? How many 5-bit strings have weight 1? How many 5-bit strings have weight 2? govt select committees

Recall that a 6-bit string is a bit strings of length 6, and a …

Web6.3 pg 413 # 11 How many bit strings of length 10 contain a)exactly four 1s? This is just asking us to choose 4 out of 10 slots to place 1’s in. C(10;4) = 10!=(4! 6!) = (10 9 8 7)=4! = 210. b)at most four 1s? We add up the number of bit strings of length 10 that contain zero 1s, one 1, two 1s, three WebQuestion: (1 point) How many 7-bit strings (that is, bit strings of length 7) are there which: 1. Start with the sub-string 101 ? 2. Have weight 5 (i.e., contain exactly five 1 's) and start with the sub-string 101 ? 3. Either start with 101 or end with 11 (or both)? 4. Have weight 5 and either start with 101 or end with \( 11 ? WebBoth B 1 3 and B 2 3 contain 3 bit strings: we must pick one of the three bits to be a 1 (three ways to do that) or one of the three bits to be a 0 (three ways to do that). Also, B 3 3 … children\\u0027s literacy initiative