Combinations of winning scholarships If six students are eligible for two scholarships worth 1k each, how many different combinations of 2 students winning the 2 scholarships are possible?
My attempt 6 nCr 2.
How am I wrong?
 A: If the scholarships are identical, and a student cannot win both, then your answer of $\binom{6}{2}=15$ is perfectly correct. 
If a student is allowed to win both, we have to add $6$ to this.
If the scholarships are named (highest standing in Math, highest standing in Physics) then the Math scholarship can be awarded in $6$ ways, and for each way so can the Physics scholarship, for a total of $6^2$.  
Remark: It unfortunately happens fairly often in elementary combinatorics/probability  that a "word problem" can be interpreted in several inequivalent ways. Your interpretation (two different students are chosen, the scholarships are identical) is I think the most reasonable one, but it is not the only one. 
A: You have to make sure you don't double count.
Call each student A, B, C, etc. Student A winning scholarship 1 and B winning scholarship 2 is the same event as A winning scholarship 2 and B winning scholarship 1.
A: The "choose" function $\frac{n!}{k!(n - k)!}$ ignores order. So if both of your scholarships are the same, 6 choose 2 is correct. That's the situation where Abe winning scholarship 1 and Betty winning scholarship 2 is the same as Betty winning scholarship 1 and Abe winning scholarship 2.
If that's not what you're trying to count, if you do care about order, then the "permute" function $\frac{n!}{k!}$ is what you want.
