Help please

If the three youngest students in the class are triplets, then I think the problem is implying we can assume they are identical. If that is so, we can group those 3 into "1 person" or 1 variable, named "a". The other 3 students will be "b, c, d". Now it is stars and bars and we can do how to distribute 13 stickers among 4 people. a + b + c +d = 13, and a, b, c, and d are nonnegative, we do (13 + 4 - 1 choose 4 - 1)

The answer is 16 choose 3 or 560.

:)

Its not 560. the hint is to deal with the restriction first.

Here's my approach: 

Each of the triplets receives 0 stickers: Then there are 13 stars and 2 bars so there ${15 \choose 2} = 105$ ways to distribute the stickers.

Each of the triplets receives 1 sticker:   Then there are 10 stars and 2 bars for ${12 \choose 2} = 66$ ways to distribute the stickers.

Each of the triplets receives 2 stickers: Then there are 7 stars and 2 bars for ${9 \choose 2} = 36$ ways to distribute the stickers.

Each of the triplets receives 3 stickers: Then there are 4 stars and 2 bars for ${6 \choose 2} = 15$ ways to distribute the stickers. 

Each of the triplets receives 4 stickers: Then there are 1 star and 2 bars for ${3 \choose 2} = 3$ ways to distribute the stickers. 

So, there are $105 + 66 + 36 + 15 + 3 = \color{brown}\boxed{225}$ ways to distribute the stickers.

I agree with BiulderBoi.