Question

Let \(S=\{1,2,3,4,5,6\}\) and \(X\) be the set of all relations \(R\) from \(S\) to \(S\) that satisfy both the following properties:

i. \(R\) has exactly 6 elements.

ii. For each \((a, b) \in R\), we have \(|a-b| \geq 2\).

Let \(Y=\{R \in X\) : The range of \(R\) has exactly one element\(\}\) and

\(Z=\{R \in X: R\) is a function from \(S\) to \(S\}\).

Let \(n(A)\) denote the number of elements in a set \(A\).

(1) If \(n(X)={ }^{m} C_{6}\), then the value of \(m\) is ______.

(2) If the value of \(n(Y) + n(Z)\) is \(k^2\), then \(|k|\) is ______.

Answer 1

(1) 20

\(S=\{1,2,3,4,5,6\} \quad R: S \rightarrow S\)

Number of elements in \(R=6\)

and for each \((a, b) \in R ;|a-b| \geq 2\)

\(X \rightarrow\) set of all relation \(R: S \rightarrow S\)

If \(a=1, b=3,4,5,6 \rightarrow(4)\)

\(a=2, b=4,5,6 \rightarrow (3)\)

\(a=3, b=1,5,6 \rightarrow(3)\)

\(a=4, b=1,2,6 \rightarrow (3)\)

\(a=5, b=1,2,3 \rightarrow (3)\)

\(a=6, b=1,2,3,4 \rightarrow (4) \)

Total number of ordered pairs \((a, b)\) 

\(s.t. |a-b| \geq 2\)

\(=20\)

\(\therefore n(X)\) = number of elements in X \(= {}^{20}C_6\)

\(\therefore m=20\)

(2) 36

\(S=\{1,2,3,4,5,6\} \quad R: S \rightarrow S\)

Number of elements in \(R=6\)

and for each \((a, b) \in R ;|a-b| \geq 2\)

\(X \rightarrow\) set of all relation \(R: S \rightarrow S\)

If \(a=1, b=3,4,5,6 \rightarrow(4)\)

\(a=2, b=4,5,6 \rightarrow (3)\)

\(a=3, b=1,5,6 \rightarrow(3)\)

\(a=4, b=1,2,6 \rightarrow (3)\)

\(a=5, b=1,2,3 \rightarrow (3)\)

\(a=6, b=1,2,3,4 \rightarrow (4) \)

Total number of ordered pairs \((a, b)\) 

\(s.t. |a-b| \geq 2\)

\(=20\)

\(\therefore n(X)\) = number of elements in X \(= {}^{20}C_6\)

\(\therefore m=20\)

\(Y=\) {\(R\in X\): The range of R has exactly one element}

From above, if range of R has exactly one element, then maximum number of elements in R will be 4.

\(\therefore n(Y) = 0\)

\(Z=\) {\(R\in X :R\) is a function from \(S\) to \(S\)}

\(n(Z) = {}^4 C_1 \times {}^3C_1 \times {}^3C_1 \times {}^3C_1 \times {}^3C_1 \times {}^4C_1\)

\(= (36)^2\)

\(n(y) + n(z) = 0 + (36)^2 = k^2\)

\(\Rightarrow |k| = 36\)