To solve the problem of counting how many bit strings of length 10 contain at least three 1s and at least three 0s, we will follow these steps:
1. Calculate Total Bit Strings
The total number of bit strings of length 10 can be calculated as:
- Total bit strings = 210 = 1024
2. Identify Available Cases
Next, we need to find the cases of bit strings that we do not want: those with less than three 1s and less than three 0s.
- Case 1: Less than 3 1s (i.e., 0, 1, or 2 1s)
- Case 2: Less than 3 0s (i.e., 0, 1, or 2 0s)
3. Calculate Unwanted Cases
Now let’s compute the number of unwanted cases for each of these cases:
Case 1: Less than 3 1s
- 0 ones: 1 way (0000000000)
- 1 one:
- Choose 1 position for the 1 in 10: C(10, 1) = 10
- 2 ones:
- Choose 2 positions for the 1s in 10: C(10, 2) = 45
So, total for Case 1 = 1 + 10 + 45 = 56 ways.
Case 2: Less than 3 0s
- 0 zeros: 1 way (1111111111)
- 1 zero:
- Choose 1 position for the 0 in 10: C(10, 1) = 10
- 2 zeros:
- Choose 2 positions for the 0s in 10: C(10, 2) = 45
So, total for Case 2 = 1 + 10 + 45 = 56 ways.
4. Avoid Double-Count
Since the case of 3 or more 1s implies 7 or more 0s and vice versa, we need to ensure not to double-count bit strings that either fit the zero or one conditions at the same time. When analyzing this, we will find that in the overlap between Case 1 and Case 2 where their sums don’t interact at the boundary instance (meaning there can’t be sequences of binary bits satisfying both simultaneously).
5. Total Valid Cases
Finally, valid cases are computed by:
- Total strings – Unwanted cases (for both cases, as no overlap):
Valid bit strings = 1024 – (56 + 56) = 1024 – 112 = 912
Conclusion
Thus, the total number of bit strings of length 10 that contain at least three 1s and at least three 0s is 912.