Question

Consider a binary code with 12 bits in each code sequence.

a. How many different sequences are possible?

b. How many sequences have at most 9 zeroes?

c. A student invented a new transistor that allows a new bit to exist with three states (0, 1, or 2) in place of two. Solve parts a. and b. for this new bit

Answer 1

a. The number of different sequences with 12 bits is 4096. b. The number of sequences with at most 9 zeroes is 286. c. For a new bit with three states, the number of sequences is 531441 with 12 bits.

Step-by-step explanation:

a. To determine the number of different sequences possible with 12 bits, we can use the concept of permutations. Since each bit can have 2 possible values (0 or 1), there are 2 options for the first bit, 2 options for the second bit, and so on.

Therefore, the total number of different sequences is 2^12 = 4096.

b. To find the number of sequences with at most 9 zeroes, we can consider the number of sequences with exactly 9 zeroes and the number of sequences with exactly 10 zeroes.

The number of sequences with exactly 9 zeroes is given by the binomial coefficient C(12, 9) = 220.

Similarly, the number of sequences with exactly 10 zeroes is C(12, 10) = 66.

Adding these two numbers gives us the total number of sequences with at most 9 zeroes: 220 + 66 = 286.

c. For the new bit with three states (0, 1, or 2), the number of different sequences can be calculated in a similar manner as in part a.

Since each bit now has 3 possible values, the total number of different sequences with 12 bits is 3^12 = 531441.

Similarly, the number of sequences with at most 9 zeroes can be calculated by considering the number of sequences with exactly 9 zeroes and 10 zeroes, as in part b.