Please help as soon as possible

44

The only choice to be made is how many gummy bears to give each child; once we've decided that, each child must receive the unique number of candies that will bring their total to 3 items. Thus, our problem reduces to counting integer solutions to a+b+c+d=6 that satisfy 0 <= a, b, c, d <= 3.

Ignoring the <= 3 restrictions, we have 9C3 = 84 solutions (which correspond to arrangements of 6 objects and 3 dividers).

Now we correct for the overcount. We must reject 5C3 = 10 solutions satisfying a >= 4, as well as 10 solutions with b, c, d >= 4. No further correction is needed, as it is not possible for more than one of a, b, c, d to be greater than or equal to 4.

Thus there are 84 - 4 x 10 = 44 ways to distribute the gummy bears, each of which forces a unique distribution of candies. There are 44 solutions.

But you guys should USE THE MESSAGE BOARD as well if you have trouble with AoPS HW problems.

(Solution from AoPS)

Hope that helped,

-sirswagger21