The number of eight-bit computer words formed by 0s and 1s with exactly three 1s and five 0s is:

To find the number of eight-bit computer words with exactly three 1s and five 0s, you can use combinations. 

An eight-bit computer word has a total of 8 positions, and you want to place three 1s in these positions. Here's the step-by-step solution:

Step 1: Choose the positions for the 1s
⇒ You want to choose 3 positions out of 8 to place the 1s.

This is a combination problem, and you can calculate it using the binomial coefficient: 

\(\ C(n, k) = \frac{n!}{k!(n-k)!} \ \)

In this case, \( C(8, 3)\) represents choosing 3 positions out of 8.

\(\ C(8, 3) = \frac{8!}{3!(8-3)!} \)

\(\ C(8, 3) = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} \)

\(\ C(8, 3) = 56 \ \)

So, there are 56 ways to choose the positions for the 1s.

Step 2: Fill the chosen positions with 1s
⇒ Once you have chosen the positions, you need to fill them with 1s. There are no choices here; you just place 1s in the positions you selected.

Step 3: Fill the remaining positions with 0s
⇒ You have 5 remaining positions to fill with 0s. Again, there are no choices here; you just place 0s in the remaining positions.

So, the total number of eight-bit computer words with exactly three 1s and five 0s is the product of the results from Step 1, Step 2, and Step 3:

\(\ \text{Total ways} = 56 \times 1 \times 1 = 56 \)

Therefore, there are 56 such eight-bit computer words.