We throw five dice of different colors. In how many ways can the sum of the throws be 11?
First, we show that 11 identical items can be placed in 5 different locations in $\binom{10}{4}$ ways such that each location gets at least one. We lay out the items in a row, numbering them from 1 to 11. Then we select which ones to place in the first location: since the items are identical, we can assume these are the first $k$ items, where $k=1,2, \ldots$. After placing the last item in the first location, we place a marker, and we proceed similarly for the other locations. This way, we associate each placement with the placement of 4 markers among the 11 items, where markers cannot be placed at the beginning or end of the row, and no more than one marker can be placed in the same spot. On the other hand, a placement of items can be reconstructed from a combination of markers: the items to the left of the first marker go to the first location, those between the first and second markers go to the second location, and so on, with those to the right of the fourth marker going to the fifth location. This way, each location gets at least one item. Since the 4 markers can be placed in 10 different spots (after the first item, after the second, ..., after the tenth), the number of marker combinations, and thus the number of possible item placements, is $\binom{10}{4}=\frac{10 \cdot 9 \cdot 8 \cdot 7}{4 \cdot 3 \cdot 2 \cdot 1}=210$.

This almost matches the number of dice rolls in question. The difference is that out of the 11 dots to be placed on 5 dice, no more than 6 can go to the same die. This modifies the number of all possible placements as follows. The possible incorrect placements (i.e., those that cannot be dice rolls) are those in which at least 7 dots go to one place. However, since at least 4 dots also go to the other 4 places, and there are 11 dots in total, this can only happen if the distribution is $7 ; 1 ; 1 ; 1 ; 1$ in some order. The number of such cases is 5, so the number of valid dice rolls is $210-5=205$.