How many bit strings of length 10 contain at least three 1s and at least three 0s?

To solve this problem, we need to count the total number of bit strings of length 10 that meet the criteria of containing at least three 1s and at least three 0s.

First, let’s determine the total number of bit strings of length 10. Each bit can either be a 0 or a 1, so there are a total of:

2¹⁰ = 1024

Now, we need to remove the bit strings that do not meet our criteria. This includes the strings that contain less than three 1s (0, 1, or 2 ones) and those that contain less than three 0s (0, 1, or 2 zeros).

To do this, we can use the concept of combinations. Let’s find the number of valid strings that we need to exclude.

1. Strings with fewer than 3 ones:

• 0 ones: There is 1 such string (0000000000).
• 1 one: The number of such strings is given by choosing 1 position from 10 for the 1, which is C(10,1) = 10.
• 2 ones: The number of such strings is given by choosing 2 positions from 10 for the 1s, which is C(10,2) = 45.

Thus, the total number of strings with fewer than 3 ones is:

1 + 10 + 45 = 56

2. Strings with fewer than 3 zeros:

• 0 zeros: There is 1 such string (1111111111).
• 1 zero: The number of such strings is given by choosing 1 position from 10 for the 0, which is C(10,1) = 10.
• 2 zeros: The number of such strings is given by choosing 2 positions from 10 for the 0s, which is C(10,2) = 45.

Thus, the total number of strings with fewer than 3 zeros is:

1 + 10 + 45 = 56

3. Overlap: Strings with fewer than 3 ones AND fewer than 3 zeros:

• These are the strings that are either all 0s or all 1s, as they cannot have 3 of either. That results in just 2 strings.

According to the principle of inclusion-exclusion, the total count of strings that do not meet the criteria (having fewer than 3 ones or fewer than 3 zeros) is:

56 (fewer than 3 ones) + 56 (fewer than 3 zeros) – 2 (overlap) = 110

Finally, to find the number of valid bit strings of length 10 that have at least three 1s and at least three 0s, we subtract the invalid strings from the total number of strings:

1024 – 110 = 914

Therefore, the number of bit strings of length 10 that contain at least three 1s and at least three 0s is 914.