Let $\bigcup_{i=1}^{50} X_i = \bigcup_{i=1}^{n} Y_i = T$ where each $X_i$ contains $10$ elements and each $Y_i$ contains $5$ elements. If each element of the set $T$ is an element of exactly $20$ of sets $X_i$'s and exactly $6$ of sets $Y_i$'s, then $n$ is equal to:

Let the relations $R_1$ and $R_2$ on the set $X = \{1, 2, 3, ..., 20\}$ be given by $R_1 = \{(x, y) : 2x - 3y = 2\}$ and $R_2 = \{(x, y) : -5x + 4y = 0\}$. If $M$ and $N$ be the minimum number of elements required to be added in $R_1$ and $R_2$, respectively, in order to make the relations symmetric, then $M + N$ equals

Let R be a relation defined on $N$ as $aRb$ if $2a + 3b$ is a multiple of $5$, $a, b \in N$. Then R is

Consider the relations $R_1$ and $R_2$ defined as $aR_1b \Leftrightarrow a^2 + b^2 = 1$ for all $a, b \in R$ and $(a, b)R_2(c, d) \Leftrightarrow a+ d = b + c$ for all $(a, b), (c, d) \in N \times N$. Then:

Let $S = {1, 2, 3, …, 10}$. Suppose $M$ is the set of all the subsets of $S$, then the relation $R = {(A, B) : A ∩ B ≠ 𝜙; A, B ∈ M}$ is :