To solve this problem, we need to consider two separate cases: bit strings that begin with three 0s and bit strings that end with two 0s.

1. Bit strings of length 10 that begin with three 0s:
Here, the first three bits are fixed as 0s. So, we only need to consider the remaining 7 bits. Each of these 7 bits can be either a 0 or a 1, giving us 2 options for each bit. Therefore, there are 2^7 = 128 such bit strings.

2. Bit strings of length 10 that end with two 0s:
In this case, the last two bits are fixed as 0s. So, we only need to consider the first 8 bits. Each of these 8 bits can be either a 0 or a 1, giving us 2 options for each bit. Therefore, there are 2^8 = 256 such bit strings.

However, if we simply add these two results together, we will be double-counting the bit strings that both begin with three 0s and end with two 0s. These bit strings have three 0s at the beginning and two 0s at the end, leaving 5 bits that can be either a 0 or a 1. Therefore, there are 2^5 = 32 such bit strings.

So, the total number of bit strings of length 10 that either begin with three 0s or end with two 0s is 128 + 256 - 32 = 352.

Question

To solve this problem, we need to consider two separate cases: bit strings that begin with three 0s and bit strings that end with two 0s.

1. Bit strings of length 10 that begin with three 0s: 
   Here, the first three bits are fixed as 0s. So, we only need to consider the remaining 7 bits. Each of these 7 bits can be either a 0 or a 1, giving us 2 options for each bit. Therefore, there are 2^7 = 128 such bit strings.

2. Bit strings of length 10 that end with two 0s: 
   In this case, the last two bits are fixed as 0s. So, we only need to consider the first 8 bits. Each of these 8 bits can be either a 0 or a 1, giving us 2 options for each bit. Therefore, there are 2^8 = 256 such bit strings.

However, if we simply add these two results together, we will be double-counting the bit strings that both begin with three 0s and end with two 0s. These bit strings have three 0s at the beginning and two 0s at the end, leaving 5 bits that can be either a 0 or a 1. Therefore, there are 2^5 = 32 such bit strings.

So, the total number of bit strings of length 10 that either begin with three 0s or end with two 0s is 128 + 256 - 32 = 352.

Knowee AI · Accepted Answer

To solve this problem, we need to consider two separate cases: bit strings that begin with three 0s and bit strings that end with two 0s.

1. Bit strings of length 10 that begin with three 0s: 
   Here, the first three bits are fixed as 0s. So, we only need to consider the remaining 7 bits. Each of these 7 bits can be either a 0 or a 1, giving us 2 options for each bit. Therefore, there are 2^7 = 128 such bit strings.

2. Bit strings of length 10 that end with two 0s: 
   In this case, the last two bits are fixed as 0s. So, we only need to consider the first 8 bits. Each of these 8 bits can be either a 0 or a 1, giving us 2 options for each bit. Therefore, there are 2^8 = 256 such bit strings.

However, if we simply add these two results together, we will be double-counting the bit strings that both begin with three 0s and end with two 0s. These bit strings have three 0s at the beginning and two 0s at the end, leaving 5 bits that can be either a 0 or a 1. Therefore, there are 2^5 = 32 such bit strings.

So, the total number of bit strings of length 10 that either begin with three 0s or end with two 0s is 128 + 256 - 32 = 352.

How many bit strings of length 10 either begin with three 0s or end with two 0s

Question

Solution

Similar Questions

Upgrade your grade with Knowee