I am trying to prove or disprove the following Lemma:
Let $S=[n]$ and $\mathcal{T}$ be the set of all $k$-subsets of $S$ that contain $t \in [n]$. Furthermore, let $\mathcal{R}$ be the set of all $k$-subsets of $S$ that do no contain $t$. Then, by randomly picking $|\mathcal{R}|$ elements from $\mathcal{T}$ denoted as $T_1, \dots, T_{|\mathcal{R}|}$, there exists an ordered tuple $u = (u_1, \dots, u_{|\mathcal{R}|})$ with $u_i \in S \setminus T_i$ for all $i \in [|\mathcal{R}|]$ such that $\bigcup_{i=1}^{|\mathcal{R}|} T_i \setminus \{t\} \cup \{u_i\} = \mathcal{R}$.
Is it possible to choose $|\mathcal{R}|$ elements from $\mathcal{T}$ denoted as $T_1, \dots, T_{|\mathcal{R}|}$ and an ordered tuple $u = (u_1, \dots, u_{|\mathcal{R}|})$ with $u_i \in S \setminus T_i$ for all $i \in [|\mathcal{R}|]$ such that $\bigcup_{i=1}^{|\mathcal{R}|} \{(T_i \setminus \{t\}) \cup \{u_i\} \} = \mathcal{R}$.
I don't know whether or not it is true but after trying some examples it seems to be true, e.g., for $n=5$, $k=3$ and $t=3$. I can recover $\mathcal{R}$ from $\{1, 3, 4\}$, $\{1, 3, 5\}$, $\{2, 3, 4\}$, $\{1, 3, 5\}$ by picking the sequence $(2,4,5,1)$$(2,4,5,2)$. Would it be hard to formally prove this?
