Candy
ANguyen
How many ways can three different pieces of candy be given to five people? A single person may get all three pieces of candy. I think its 5 pick 3 but the fact that one person may get all three pieces is throwing me off. Please help
rachelmernoff: Jan. 27, 2015, 9:08 p.m.
Pretend that each candy is a position in a line, except that when one person takes one position, they can also take another position. Therefore, for the first piece of candy, there are 5 possible outcomes (since there are 5 different people the candy can be given to). For the second and third pieces of candy, there are also 5 possible outcomes for each because a person can receive more than one piece. The answer then is 5^3.
ANguyen: Jan. 28, 2015, 4:35 p.m.
The answer is actually 5x4x3+5x(3,2)x4+5 but I have no idea why
NatalieNguyen2F: Jan. 29, 2015, 4:48 p.m.
Both 5^3 and 5x4x3 + 5x(3, 2)x4 + 5 equal 125 ways. The second answer just uses different logic. The second answer splits the problem into three cases: CASE 1 is when no one gets more than 1 candy, CASE 2 is when one person gets exactly 2 candies, and CASE three is when one person gets all 3. In CASE 1, you must first choose the 3 out of 5 people and multiply it by the number of ways to give the candy: C(5,3) x 3! (also the same as 5x4x3). In CASE 2, you multiply the number of ways to choose the guy to gets the 2 candies, the way to choose the 2 candies, and the ways to pick the last guy from the 4 left: 5 x C(3,2) x 4. In CASE 1, you just write the number of ways to pick the one guy that gets all the candy: 5. You finish by adding all these different cases together. Personally, I think thinking about it the way Rachel suggested is much simpler.
weisbart: Feb. 1, 2015, 11:02 a.m.
You guys are great. Perfect answers. rachelmernoff's reasoning is right on and NatalieNguyen2F's explanation is exactly correct. I like the $5^3$ reasoning as well much better. It's always best to avoid cases.