Problem 5 (Partition problem) In how many ways can 11 items be divided into two groups such that each group has no fewer than three items (selecting one group uniquely determines the other)?
Solution. All possibilities to divide 11 items into two groups are represented by the following decompositions: $11=3+8, 11=4+7$, $11=5+6$ (the possibilities $11=6+5, 11=7+4$ and $11=8+3$ coincide with the previous ones). It remains to understand in how many ways one can choose 3, 4, or 5 items out of 11. Since the order of selection does not matter, we are dealing with combinations: the corresponding number of ways are $C_{11}^{3}$, $C_{11}^{4}$, and $C_{11}^{5}$. Under these conditions, the addition rule applies to three objects. Therefore, dividing 11 items into two groups so that each group has at least three items is possible in $C_{11}^{3}+C_{11}^{4}+C_{11}^{5}=957$ ways.

Answer. 957