1. My favorite type of problem! Get ready for Stars and Bars! To read more about "Stars and Bars," check here: https://math.stackexchange.com/questions/910809/how-to-use-stars-and-bars
Anyway, when we plug \(n=6\) and \(k=3\)\(\), we have \(\binom{6+3-1}{3-1}=\binom{8}{2}=4*7=\boxed{28}.\)
2. We break this up into cases. To create 6, with three numbers, we can use: 6-0-0, 5-1-0, 4-2-0, 4-1-1, 3-3-0, 3-2-1, and. 2-2-2.
6-0-0: Only one way
5-1-0: Take the ball that is alone, so 6 ways.
4-2-0: 6C2=15 ways for the two balls in their own box.
4-1-1: Again 15 ways, pick two of the six balls, but they cannot go in the box of 4.
3-3-0: 5C2=10 ways, by selecting two of the remaining five balls.
3-2-1: 6*10=60 ways by picking any of the six balls to be by itself.
2-2-2: 15 ways? First, you think they are distinguishable, yet you overcount by a factor 3!=6.
Thus, adding them all up, we get \(1 + 6 + 15 + 15 + 10 + 60 + 15 = \boxed{122}. \)
3. Try this on your own! Hint: Just multiply by the same number of outcomes of the white balls for the black balls!
