To solve this problem, we need to determine how many different 8-bit binary strings can be formed that contain exactly five 0's and three 1's.

Step-by-step explanation:

1. Total positions in the binary string:
   Since we're dealing with an 8-bit binary string, there are 8 positions in total.

2. Choosing positions for 1's (or 0's):
   Out of these 8 positions, we need to place exactly three 1's, and the rest will be 0's (which means five 0's). The problem of finding the number of such strings is now reduced to choosing 3 positions for the 1's from the 8 positions available.

3. Using combinations:
   The number of ways to choose 3 positions out of 8 is given by the binomial coefficient:

   $\binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!}$

4. Calculating the value of the combination:

   $\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56$

Therefore, the number of binary strings that contain exactly five 0's and three 1's is 56.

Counting argument:

• The essence of the counting argument is that once we choose 3 positions to place the 1's, the remaining positions will automatically be filled with 0's. Since the order of choosing the 1's does not matter, we use combinations rather than permutations. This ensures that we count only unique arrangements.

Would you like more details or have any questions about this?

Here are 5 related questions:

1. What is the number of 8-bit binary strings containing exactly four 1's and four 0's?
2. How many 10-bit binary strings contain exactly six 1's and four 0's?
3. What is the total number of 8-bit binary strings (without any restrictions)?
4. How can you generalize this counting method for any $n$-bit string with $k$ 1's?
5. How do binomial coefficients relate to Pascal's triangle?

Tip: When solving problems involving binary strings and choosing specific positions, combinations (binomial coefficients) are a key tool for determining the number of valid arrangements.