Final answer:
To find bit strings of length 12 with at most three 1's, we use combinations to calculate the sum of strings with 0, 1, 2, and 3 ones, which gives us 299. To find bit strings with at least three 1's, we subtract the bit strings with 0, 1, and 2 ones from the total number of bit strings, leading to 4017 bit strings.
Step-by-step explanation:
To solve the problem of counting bit strings of length 12 with certain constraints, we need to use combinatorics. Specifically, we use combinations to determine how many ways we can choose positions for the 1's in the string.
a) At most three 1's:
To find the number of bit strings of length 12 that contain at most three 1's, we need to consider all possibilities with 0, 1, 2, and 3 ones:
- With 0 ones: C(12, 0) = 1 way
- With 1 one: C(12, 1) = 12 ways
- With 2 ones: C(12, 2) = 66 ways
- With 3 ones: C(12, 3) = 220 ways
Summing these up gives us 1 + 12 + 66 + 220 = 299 bit strings.
b) At least three 1's:
The number of bit strings of length 12 that contain at least three 1's can be found by subtracting the number of strings with 0, 1, and 2 ones from the total number of bit strings:
Total number of bit strings of length 12 is 212 = 4096.
Strings with 0, 1, or 2 ones (as calculated before): 1 + 12 + 66 = 79.
Therefore, the number of strings with at least three 1's = 4096 - 79 = 4017 bit strings.