Question

A team of 4 students is chosen every year to attend a meeting.
a) How many ways can be the team chosen if there are 12 eligible students?
b) How many ways, if two of the students are brothers and will only attend if both do?
c) How many ways, if two of the students can't attend at the same time?

Answer 1

\[a: \frac{12!}{4!}{(12-4)!}\]
For b count the number of ways with zero brothers (pick 4 from 10) and add that to the number of ways to pick the brothers plus 2 from 10
For c consider three cases:
1) neither of the brats attend pick 4 from 10
2) brat a attends but not b :: pick 3 from 10
3) brat b attends but not a :: pick 3 from 10

Answer 2

Silly LaTeX, I mean:
\[\frac{12!}{4!(12-4)!}\]