How many bit strings of length 10 both begin and end with a 1?
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In a bit string length 12 How many begin with 110?You are essentially asking how many different bit strings of length 9 are there, as the first 3 bits are fixed. The answer is 2^9 = 512. How many bit strings of length 8 are there which begin with a 0 and end with a 0?-- There are 256 bit strings of length 8 . -- There are 4 bit strings of length 2, and you've restricted 2 of the 8 bits to 1 of those 4 . -- So you've restricted the whole byte to 1/4 of its possible values = 64 of them. How many different bit strings are there of length 7?there are 128 (2 to the power of 7) bit strings of length 7 How many bit strings of length 8 contain either three consecutive 0s or four consecutive 1s?Five (5) have one or the other but not both. Six (6) have both. Total of eleven (11). How many bit strings of length not exceeding n consist entirely of 1s?n+1 (counting the empty string) Video TranscriptHello there. So for this exercise we need to determine how many beat the strength of Langton uh, access that begin with and now an end with the number one. So technically you have here, the spring of lengthen. Okay, so you have here some other places. The point is that here the last number is going to be one and the initial number is going to be one as well. So at the end you see that the Bits that are going to be in the middle of the strength are going to be eight places. And this each place here can have either a zero or a one because it's a bit strength. So we have two possibilities for each of these places in the B string. So we have two times to the times two And this repeats eight times. So this result into two to the 8th power And the result is 256 uh possible be the strength of linton that started with one and end with one $\begingroup$
My answer would be: $2^6$. Because we know, that the bit starts with $1$ and ends with $1$: Which means, that $2$ of the length $8$ is used, also there are $6$ positions remaining, which is $2^6 = 64$. Is this the correct answer?
Frentos 2,9011 gold badge9 silver badges22 bronze badges asked Feb 6, 2016 at 23:56
$\endgroup$ 3 $\begingroup$ Yes the answer $2^6$ is correct. Two bits you know are $1$ already, so there are $8-2=6$ bits left. Each bit can either be a $0$ or a $1$, so you have $2$ possibilities for each slot. Hence, as you concluded correctly, you have $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 2^6$ possible combinations for your string.
answered Feb 7, 2016 at 0:08 $\endgroup$ 1 $\begingroup$ Since there is one choice for the first position, one choice for the last position, and two choices for each of the six positions between them, the number of bit strings of length eight that begin and end with a $1$ is $2^6$, so your solution is correct. $\endgroup$ How many bits is a string of length 10?Solution: From the question it is given that strings of length 10 contain at least three 1s and at least three 0s. The total length is = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024. 1024 - 1 - 1 - 10 - 10 - 45 - 45 = 912.
How many bit strings of length eight both begin and end with a 1?How many 8-bit strings begin and end with 1? Answer : 26 since first and last bit have been already determined.
How many bit strings of length 10 have the same number of 0s and 1s?Strings “0011”, “0101”, “0110”, “1100”, “1010” and “1001” have same frequency of 0s and 1s. Hence, the answer is 6.
How many bit strings of length seven either start with a 1 bit or end with the bit 1?The correct answer is thus 44.
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