Mikulás distributes 53 candies into three bags, making sure that each bag contains a different number of candies and that any two bags together contain more than the third one. In how many ways can he do this?
Solution. Let's denote the bags with $A$, $B$, and $C$. We place these labels on them, as otherwise the bags cannot be distinguished. Divide the 53 candies. Since their number is odd, this can only be done so that one half consists of 27 candies, and the other half of 26 candies. Place the 26 candies in bag $A$, then we need to distribute 27 candies between bags $B$ and $C$ while adhering to the given conditions $(A=26, B+C=27)$. There are as many ways to do this as there are ways to break down 27 into the sum of two numbers greater than 2 and less than or equal to 13. This is a total of: $25+2, 24+3, 23+4, \ldots, 14+13, 12$ possibilities. In each case, it holds that $A+B>C$, $A+C>B$, since $A$ is the largest, and $B+C>A$.

Now, place 25 candies in bag $A$. Distribute the remaining 28 candies between $B$ and $C$. Again, it holds that $B+C>A$. In $B$ and $C$, $(A=25, B+C=28)$, 24+4, 23+5, 22+6, ..., 15+13 candies will be placed, which is 10 possibilities.

We continue this until 19 candies are placed in $A$ and 34 candies are placed in $B$ and $C$ together. We cannot continue further because $A=18$, $B=19$, $C=16$ has already occurred, regardless of the order, and the same applies to the others.

Now, let's count how many decompositions are possible if $A=24, 23, 22, 21, 20$, or 19.

If $A=24, B+C=29$, the possible decompositions are: $23+6, 22+7, 21+8, \ldots, 15+14$, which is 9 cases;

For $A=23$, $B+C=30$, and the decompositions are: $22+8, 23+9, \ldots, 16+14$: this is another 7 possibilities;

For $A=22$, $B+C=31$, the sums are: $21+10, 20+11, \ldots, 16+15$: this is 6 possibilities;

For $A=21$, $B+C=32$, the sums are: $20+12, 21+13, \ldots, 17+15$: 4 possibilities;

For $A=20$, $B+C=33$, the sums are: $19+14, 18+15, 17+16$: 3 cases;

For $A=19$, $B+C=34$, the sum is $18+16$, which is 1 possibility.

In total: $12+10+9+7+6+4+3+1=52$. The total number of possibilities is 52, and if the bags could also be distinguished, the total number of possibilities would be $52 \cdot 3!=312$.