Answer:
a) Therefore, the minimum number of bits required is 6.
b) Therefore, we can admit 4 more students to the class without requiring additional bit for each student's unique bit pattern
Explanation:
a) The number of unique bit patterns using <em>n</em> bits is calculated using
.
In this case, there are 60 students, so, we need at least 60 unique bit pattern.
![2^{n} = 60](https://tex.z-dn.net/?f=2%5E%7Bn%7D%20%20%3D%2060)
Where <em>n</em> is the number of bit required; we are to find <em>n</em>
We take the logarithm of both side:
![log_{2} 2^{n} = log_{2} 60\\n = log_{2} 60\\n = 6 (Approximately)](https://tex.z-dn.net/?f=log_%7B2%7D%202%5E%7Bn%7D%20%20%3D%20log_%7B2%7D%2060%5C%5Cn%20%3D%20log_%7B2%7D%2060%5C%5Cn%20%3D%206%20%28Approximately%29)
Therefore, the minimum number of bits required is 6
b) How many more students can be admitted to the class without requiring additional bits for each student's unique bit pattern?
With 6 bits, we can represent up to
unique bit pattern which is 64 unique bit patterns.
To get the number of additional bit:
64 - 60 = 4
Therefore, we can admit 4 more students to the class without requiring additional bit for each student's unique bit pattern