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. 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

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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? WebMay 3, 2024 · Or, for maybe slightly less computation, you could say "at least 3" means not ( 0, 1, or 2 ), so (since there are 2 7 bit-strings of length 7 in all) 2 7 − ( ( 7 0) + ( 7 1) + ( 7 2)) = 99. Of course, if you've already calculated 64 for "at most 3 " and 35 for "exactly 3", you … can i get monk fruit without erythritol

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http://courses.ics.hawaii.edu/ReviewICS141/morea/counting/PermutationsCombinations-QA.pdf 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 … can i get monkeypox from the subway