How many ternary strings does a length 4 have?
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Patrick B. answered • 01/17/21 Tutor 4.7 (31) Math and computer tutor/teacher See tutors like this See tutors like this 3^8= 9^4= 81^2= 81*81= (80+1)(80+1) 6400 + 80 +80+1= 6561 Upvote • 1 Downvote Add comment More Report Still looking for help? Get the right answer, fast.Ask a question for free Get a free answer to a quick problem. ORFind an Online Tutor Now Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. Question: Kayla P. Algebra 7 months, 1 week ago Video Answer: Get the answer to your homework problem. Try Numerade free for 7 days AD Central European University We don’t have your requested question, but here is a suggested video that might help. Related QuestionA string that contains only Os, 1s, and 2s is called a ternary string a) Find a recurrence relation for the number of ternary strings of length n that do not contain any of the strings 10 and 20. Show how you obtain the recurrence b) What islare the initial condition(s)? c) How many ternary strings of length six do not contain any of the strings 10 and 202 DiscussionYou must be signed in to discuss. Video TranscriptThere are strings that don't contain the strings 10 or 20. zero can't happen after one or 2. Only at the beginning of the string can you do that. No zero can happen after one or two occur. Some examples of how the strings would look. In any of it? Zeros are not visible after one or 2. Let's say the last digit of every string is one or 2. We want to make strings that don't have a number. Is it right? If the last digit is one or two and we don't have 10 or 20 then we can add any string of length and not having 10 or 20. There are a sub and a negative. The strings are like this. A 7 is simply strings of length that don't have a 10. That is how we defined seven. In case one we have two times A and minus one. There was a seven and one. The last digit can be one or two, so the strings of length don't have a number. There are two choices and we have a 7 for each one. There are strings of length N-1 that can be used. Absolutely perfect. We have twice a 7 -1 together. Now is the right time to do so. We are considering what will happen when the last digit of the Lantern is zero. That only means that everything is itself. Zero cannot appear after one or two, so the string has to be zero in that case. Alkatiri's None of the digits precede the zero. One or two can be possible. They're jews. So from case one, all the digits have to be zero, and that's just one string. A seven equals seven minus one, so we have case one and case two combined. We get from case one to case two. For part b, plus one and for n greater than this reference holds. We want to figure out the first conditions. The strength of a sub one is the strength of a length that does not have 10 or 20. Length one strings are basically zero or one or two since it's one at 10 201 doctor. There are three of them, that's right. A sub two is three squared minus two and a sub one is three squared minus one. They're nowhere three squared, so we're doing this. Each position has three choices, so the turnberry strings are length two. We are removing the two that are 10 or 20 from all of the strings. Three squared minus two equals seven, so it's up to him. In part C, we are going to find out a sub six, and these are the initial conditions. The records we saw here show that a sub three would be twice a sub two and one. Sub three is twice a sub two plus one and that's up to seven plus one, which is 15, so that's two times seven. Then we find a four which is given by two times a sub three plus one and which we found was 15 in the previous steps. We have 15 plus one equals 31. Two times is a four plus one which is 31 plus one. A six would be twice a sub five plus one, which is two times 63 plus one, which gives us 27. For this one, that's it. |
