Draw the Venn diagram of a generic configuration of three subsets $A,B,C$ of $U$, of cardinalities $|A| = a$, $|B| = b$, $|C|=c$, a fourth subset $\emptyset \subseteq Z \subseteq A\cap B\cap C$ and a fifth subset $(A\cap B\cap C)\setminus Z \subseteq Y \subseteq A\cup B\cup C$, such that $Y\cap Z = \emptyset$. Denote $\bullet$ $T_{abc} = A\cap B\cap C\cap Y$, $\bullet$ $T_{ab} = (A\cap B\cap Y)\setminus T_{abc}$ and similarly $T_{bc}$, $T_{ca}$, $\bullet$ $T_{a} = (A\cap Y) \setminus (T_{ab}\cup T_{ac}\cup T_{abc})$ and similarly $T_{b}$, $T_{c}$, $\bullet$ $S_{abc} = (A\cap B\cap C)\setminus Y$, $\bullet$ $S_{ab} = (A\cap B)\setminus (Y\cup S_{abc})$ and similarly $S_{bc}$, $S_{ca}$, $\bullet$ $S_{a} = A\setminus (Y\cup S_{ab} \cup S_{bc} \cup S_{abc})$ and similarly $S_{b}$, $S_{c}$, $\bullet$ $\Omega = U \setminus (A\cup B\cup C)$. The $15$ subsets described above make up a partition of $U$, with $Z = S_{abc}$, $\displaystyle Y = \left (\bigcup_{a} T_{a}\right ) \cup \left(\bigcup_{ab} T_{ab}\right )\cup T_{abc}$, and, denoting $X = (A\cup B\cup C) \setminus (Y\cup Z)$, $\displaystyle X = \left (\bigcup_{a} S_{a}\right ) \cup \left(\bigcup_{ab} S_{ab}\right )$. Without the restriction on the cardinalities, there are $15^n$ possibilities of such configurations; with the restrictions, there will be some number $N$, which we do not care and do not need to compute. Now denote $X_{a} = A = S_{a} \cup S_{ab} \cup S_{ca} \cup S_{abc}\cup T_{a} \cup T_{ab} \cup T_{ca} \cup T_{abc}$, and similarly $X_{b}$ and $X_{c}$. If we consider a bijection $\sigma \colon \{1,2,3\} \to \{a,b,c\}$ (where $a,b,c$ are taken as letters, not the values of the cardinalities), we will have to have $\bullet$ $X_{i} = X_{\sigma(i)}$ for all $1\leq i \leq 3$; $\bullet$ $Y_{1} = T_{\sigma(1)}$, $\bullet$ $Y_{2} = T_{\sigma(2)} \cup T_{\sigma(1)\sigma(2)}$, $\bullet$ $Y_{3} = T_{\sigma(3)} \cup T_{\sigma(1)\sigma(3)}\cup T_{\sigma(2)\sigma(3)}\cup T_{\sigma(1)\sigma(2)\sigma(3)}$. Because of the obvious symmetry of notations, the number of ordered sextuplets is the same for any bijection $\sigma$, i.e. for any permutation of $a,b,c$, namely the same value $N$.

International Zhautykov Olympiad 2014, Problem 5 (Second day).

The following is, I think, a better, more to the point and more convincing approach (albeit based on the same idea). In order to avoid any confusion between the letters $a,b,c$ and their numerical values (as cardinalities of sets), the most convenient way will be to denote by $|\ell|$ the cardinality symbolized by any such letter $\ell$. We can now consider the true $3$-element set $\{a,b,c\}$, and the canonical bijection $\phi\colon \{1,2,3\} \to \{a,b,c\}$ given by $\phi(1)=a$, $\phi(2)=b$, $\phi(3)=c$. Let us now consider any permutation $\sigma$ of $\{a,b,c\}$. We will denote by $\mathcal{F}_{\sigma}$ the family of ordered sextuplets $(X_{1},X_{2},X_{3},Y_{1},Y_{2},Y_{3})$ of subsets of $U$ satisfying the conditions of the statement but, under our notations, having $|X_{1}|=|\sigma(a)|$, $|X_{2}|=|\sigma(b)|$, $|X_{3}|=|\sigma(c)|$. We will also denote by $\mathcal{F}$ the family of doubletons $\{(X_a,X_b,X_c), Y\}$, with $X_{a},X_{b},X_{c}$ subsets of $U$ having $|X_{a}|=|a|$, $|X_b|=|b|$, $|X_c|=|c|$, and $\emptyset \subseteq Y\subseteq X = X_a\cup X_b\cup X_c$. The set $Y$ uniquely partitions into $7$ classes (some of them maybe empty), indexed by the non-empty subsets $S$ of $\{a,b,c\}$ via $\displaystyle Y_S = Y\cap \left (\bigcap_{\ell \in S} X_\ell\right )\cap \left (\bigcap_{\ell \not \in S} (U\setminus X_\ell)\right )$. We will now establish a bijection between $\mathcal{F}$ and $\mathcal{F}_\sigma$. This will show that $F(\sigma(a),\sigma(b),\sigma(c)) = |\mathcal{F}|$ is constant over all permutations $\sigma$. We send an element $\{(X_a,X_b,X_c), Y\} \in \mathcal{F}$ into the sextuplet $(X_{1},X_{2},X_{3},Y_{1},Y_{2},Y_{3})$ given by, and easily verified it actually belongs to $\mathcal{F}_\sigma $, $\bullet$ $X_1 = X_{\sigma(a)}$, $X_2 = X_{\sigma(b)}$, $X_3 = X_{\sigma(c)}$ (i.e. $\hspace {-5pt}$ $X_i = X_{\sigma(\phi(i))}$ for $i\in \{1,2,3\}$), $\bullet$ $Y_1 = Y_{\{\sigma(a)\}}$, $\bullet$ $Y_2 = Y_{\{\sigma(b)\}} \cup Y_{\{\sigma(a), \sigma(b)\}}$, $\bullet$ $Y_3 = Y_{\{\sigma(c)\}} \cup Y_{\{\sigma(a), \sigma(c)\}}\cup Y_{\{\sigma(b), \sigma(c)\}}\cup Y_{\{\sigma(a),\sigma(b),\sigma(c)\}}$. We also send an element $(X_{1},X_{2},X_{3},Y_{1},Y_{2},Y_{3})\in \mathcal{F}_\sigma $ into the doubleton $\{(X_a,X_b,X_c), Y\}$ given by, and easily verified it actually belongs to $\mathcal{F}$, with $\tau$ the permutation of $\{1,2,3\}$ induced by $\sigma$ via $\tau(i) = \phi^{-1}(\sigma(\phi(i)))$ for all $i\in \{1,2,3\}$, $\bullet$ $X_a = X_{\tau(1)}$, $X_b = X_{\tau(2)}$, $X_c = X_{\tau(3)}$ (i.e. $X_{\phi(i)} = X_{\tau(i)}$ for $i\in \{1,2,3\}$), $\bullet$ $Y = Y_1\cup Y_2\cup Y_3$. It is immediate to see this mapping is a bijection, due to the unicity of the partitioning described in the above. Visualizing the Venn diagrams should tremendously help in understanding our considerations. The only difficulty resides in providing a luminous write-up of the argumentation, the underlying phenomenon being in fact almost trivial. The key element of this solution is to consider the unique partitioning of the set $Y = Y_1\cup Y_2\cup Y_3$ induced by the three sets of cardinalities $a$, $b$, $c$.

Can't we compute $f(a,b,c)$ ??

Can somebody evaluate $f(a,b,c)$? $f(a,b,c)=C_{2014}^a \cdot \sum_{i=0}^a \big{(} C_a^i \cdot C_{2014-i}^b \cdot \sum_{j=0}^b (C_b^j \cdot C_{2014-i-j}^c) \big{)}\cdot 2^c$ correct?