Question

The no. of 6 bit strings
either begins with 11or end with 101 are​

Answer 1

Answer:  As $2^{4}$ and $3^{3}$.

Step-by-step explanation: As the no. of 6-bit strings

As it either begins with 11 or ends with 101.

As the strings start with 11: 11xxxx

So $2^{4}$ Combination.
And now ends with 101: xxx101

So $3^{3}$ Combination.

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Answer 2

Answer:

The number of 6-bit strings that either begins with 11 or end with 101 is 22.

Step-by-step explanation:

We know that the number of n-bit strings possible is given as $2^{n}$.

Now, for the string that begins with 11,
Out of 6-bits, two places are already fixed. Therefore, the number of strings starting with 11 is given as,
$2^{6-2} =2^{4} =16$

Now for the string that ends with 101,
Out of 6-bits, three places are already fixed. Therefore, the number of strings ending with 101 is given as,
$2^{6-3} =2^{3} = 8$

Therefore, the total number of strings that either begins with 11 or ends with 101 $=16+8-2=22$ (two is subtracted for repeated strings considered in both cases).

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