WebCase 1: Have weight 5. From the 9 places to have digits, there are 5 places to have 1's. That's "9 choose 5" = 9C5 = 126 Case 2: Start with 101 That's problem a), or 64 Case 3. Have weight 5 and start with 101. That's problem b), or 20 Let A = the set of 9-bit strings with weight 5. Let B = the set of 9-bit strings that start with 101. WebTo count all of these strings, we must include all 4-bit strings of weight 3. In the second case (the string starts with a 1), we still have four bits to choose, but now only two of them can be 1's, so we should look at all the 4-bit strings of weight 2.
10) How many 8-bit strings contain six or more 1s? - UTEP
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 … trustpharmacy review
Answered: Recall that a 5-bit string is a bit… bartleby
Web(c) Three leftmost positions and two rightmost positions are just occupied and, THEREFORE, are not the subject of consideration. For the rest 12-3-2 = 7 positions, we have 7-2-1 = 4 "ones" to distribute. We can choose these 4 positions for "ones" from 7 positions by ways. Solved. All questions (a), (b) and (c) are answered. WebApr 1, 2024 · An 8-bit string has 8 bits (A bit is a binary digit). A bit is either a 0 or a 1. The weight of a bit string is the number of 1’s that it contains. Thus, there are four 1's and four … 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 … trustpharmacy.com