As per the title:
Consider 30 distinct computers. 6 of these computers are broken. How many ways are there of choosing
- 4 of any computer
- 4 computers with exactly 2 broken computers
- 4 computers with at most one broken computer
Progress so far:
- Use the binomial coefficient: $\begin{pmatrix} 30 \\ 4 \end{pmatrix} $
- Here I am not sure. Perhaps $\begin{pmatrix} 24 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 2 \end{pmatrix} $ ?
Can anyone give me some hints on how to think about the last two questions? Also, how would my answer to the first question be different if the computers were not distinct?
Your working for (2) is correct; out of the $6$ broken computers choose $2$, out of the $24$ non-broken ones choose another $2$, then multiply.
The same reasoning applies for (3). Either you choose $1$ broken and $3$ non-broken computers $\left(\binom{24}3\binom61\right)$ or $4$ non-broken computers $\left(\binom{24}4\binom60\right)$. You then add up these numbers of choices.
If the computers were not distinct, there would only be five ways to pick the computers. Once the number of broken computers is chosen $(0, 1, 2, 3, 4)$, the remaining computers must be non-broken.