2019 AMC 8

Solution 1 

Note: This solution uses the non-negative version for stars and bars. A solution using the positive version of stars is similar (first removing an apple from each person instead of 2).

This method uses the counting method of stars and bars (non-negative version). Since each person must have at least apples, we can remove apples from the total that need to be sorted. With the remaining apples, we can use stars and bars to determine the number of possibilities. Assume there are stars in a row, and bars, which will be placed to separate the stars into groups of . In total, there are spaces for stars spaces for bars, for a total of spaces. We can now do . This is because if we choose distinct spots for the bars to be placed, each combo of groups will be different, and all apples will add up to . We can also do this because the apples are indistinguishable. is , therefore the answer is .

Solution 2 

Consider an unordered triple where and are not necessarily distinct. Then, we will either have , , or distinguishable ways to assign , , and to Alice, Becky, and Chris. Thus, our answer will be for some nonnegative integers . Notice that we only have way to assign the numbers to Alice, Becky, and Chris when . As this only happens way (), our answer is for some . Finally, notice that this implies the answer is mod . The only answer choice that satisfies this is .