Given: a team of 5 is made up of 12 students. which means n = 12, r = 5.
Find: 1.) # of possible teams. 2.) # of possible teams when 2 students won't play together. 3.) # of possible teams when 2 students MUST play together.
My attempt:
1.) nCr = 12 C 5 = 792 2.) 12C5 - 10C5 = 540? 3.) 12C3 = 220?
1) is correct.
2) is correct.
3) It is not given in the question that the two players could be left out. Hence, there are two situations : leave both people out, or include both of them. Hence the answer is $\binom{10}{3} + \binom{10}{5} =372$. (in the first case, both are chosen, and in the second case, both are not chosen).
You have to careful while including and excluding people in your combinations. Also, maybe you could have read the question a little more carefully.
