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
Step-by-step explanation:
a) The number of unique bit patterns using n bits is calculated using
.
In this case, there are 60 students, so, we need at least 60 unique bit pattern.
![2^(n) = 60](https://img.qammunity.org/2021/formulas/computers-and-technology/high-school/oce04r8l04k5p7y6t0l1yu6yz5ah692zwj.png)
Where n is the number of bit required; we are to find n
We take the logarithm of both side:
![log_(2) 2^(n) = log_(2) 60\\n = log_(2) 60\\n = 6 (Approximately)](https://img.qammunity.org/2021/formulas/computers-and-technology/high-school/jtoc5pc0r24vfvh70nukk2uwjovomicy0p.png)
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