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 . If the value of n(Y)+n(Z) is k^2 , then |k| is \underline\hspace2cm .

Permutations and Combinations PYQ — JEE Advanced 2024 · Shift 2 · Q32 with Solution

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Correct answer: 36

We need to compute $n(Y)+n(Z),$ and then write it as $k^2$ .

1. Basic allowed ordered pairs

We have $S=\{1,2,3,4,5,6\}.$ A relation $R$ from $S$ to $S$ is a subset of $S\times S$ .

Given condition:

$R$ has exactly $6$ elements.
For every $(a,b)\in R$ , we must have $|a-b|\ge 2.$

So first, for each $a\in S$ , count how many $b\in S$ satisfy $|a-b|\ge 2$ .

For $a=1$ : allowed $b=3,4,5,6$ $\Rightarrow 4$
For $a=2$ : allowed $b=4,5,6$ $\Rightarrow 3$
For $a=3$ : allowed $b=1,5,6$ $\Rightarrow 3$
For $a=4$ : allowed $b=1,2,6$ $\Rightarrow 3$
For $a=5$ : allowed $b=1,2,3$ $\Rightarrow 3$
For $a=6$ : allowed $b=1,2,3,4$ $\Rightarrow 4$

Total allowed ordered pairs: $4+3+3+3+3+4=20.$

2. Counting $n(Y)$

Set $Y$ consists of those relations in $X$ whose range has exactly one element.

So all 6 ordered pairs in $R$ must have the same second coordinate, say $b$ . Thus $R=\{(a,b): a\in A\}$ for some subset $A\subseteq S$ with $|A|=6$ . Since $S$ itself has 6 elements, necessarily $A=S.$ So if range has exactly one element and relation has exactly 6 elements, then for every $a\in S$ , the pair $(a,b)$ must be in $R$ .

Hence for a fixed $b$ , this relation is $R_b=\{(1,b),(2,b),(3,b),(4,b),(5,b),(6,b)\}.$ This is valid only if each pair satisfies $|a-b|\ge 2$ for all $a\in S$ .

Now check whether there exists such a $b$ .

If $b=1$ , then $(1,1)$ has $|1-1|=0<2$ .
If $b=2$ , then $(2,2)$ has $0<2$ .
Similarly for every $b\in S$ , the pair $(b,b)$ appears and violates the condition.

Therefore no such relation exists.

So, $n(Y)=0.$

3. Counting $n(Z)$

Set $Z$ consists of those relations in $X$ that are functions from $S$ to $S$ .

A function from $S$ to $S$ must assign exactly one image to each element of $S$ . Since $|S|=6$ and $R$ has exactly 6 elements, this fits perfectly: for each $a\in S$ , choose exactly one $b\in S$ such that $|a-b|\ge 2$ .

Thus the number of such functions is the product of the number of choices for each input:

For $1$ : $4$ choices
For $2$ : $3$ choices
For $3$ : $3$ choices
For $4$ : $3$ choices
For $5$ : $3$ choices
For $6$ : $4$ choices

Therefore, $n(Z)=4\cdot 3\cdot 3\cdot 3\cdot 3\cdot 4=16\cdot 81=1296.$

4. Compute $n(Y)+n(Z)$

$n(Y)+n(Z)=0+1296=1296.$

Given $1296=k^2,$ we get $|k|=\sqrt{1296}=36.$

5. Final answer

$\boxed{36}$

6. Comparison with stored correct answer

Stored correct answer = $36$ .

Our derived answer also gives $36$ , so they agree.

More from Permutations and Combinations

View written solutionFree

Correct answer: 36

We need to compute $n(Y)+n(Z),$ and then write it as $k^2$ .

1. Basic allowed ordered pairs

We have $S=\{1,2,3,4,5,6\}.$ A relation $R$ from $S$ to $S$ is a subset of $S\times S$ .

Given condition:

$R$ has exactly $6$ elements.
For every $(a,b)\in R$ , we must have $|a-b|\ge 2.$

So first, for each $a\in S$ , count how many $b\in S$ satisfy $|a-b|\ge 2$ .

For $a=1$ : allowed $b=3,4,5,6$ $\Rightarrow 4$
For $a=2$ : allowed $b=4,5,6$ $\Rightarrow 3$
For $a=3$ : allowed $b=1,5,6$ $\Rightarrow 3$
For $a=4$ : allowed $b=1,2,6$ $\Rightarrow 3$
For $a=5$ : allowed $b=1,2,3$ $\Rightarrow 3$
For $a=6$ : allowed $b=1,2,3,4$ $\Rightarrow 4$

Total allowed ordered pairs: $4+3+3+3+3+4=20.$

2. Counting $n(Y)$

Set $Y$ consists of those relations in $X$ whose range has exactly one element.

Hence for a fixed $b$ , this relation is $R_b=\{(1,b),(2,b),(3,b),(4,b),(5,b),(6,b)\}.$ This is valid only if each pair satisfies $|a-b|\ge 2$ for all $a\in S$ .

Now check whether there exists such a $b$ .

If $b=1$ , then $(1,1)$ has $|1-1|=0<2$ .
If $b=2$ , then $(2,2)$ has $0<2$ .
Similarly for every $b\in S$ , the pair $(b,b)$ appears and violates the condition.

Therefore no such relation exists.

So, $n(Y)=0.$

3. Counting $n(Z)$

Set $Z$ consists of those relations in $X$ that are functions from $S$ to $S$ .

Thus the number of such functions is the product of the number of choices for each input:

For $1$ : $4$ choices
For $2$ : $3$ choices
For $3$ : $3$ choices
For $4$ : $3$ choices
For $5$ : $3$ choices
For $6$ : $4$ choices

Therefore, $n(Z)=4\cdot 3\cdot 3\cdot 3\cdot 3\cdot 4=16\cdot 81=1296.$

4. Compute $n(Y)+n(Z)$

$n(Y)+n(Z)=0+1296=1296.$

Given $1296=k^2,$ we get $|k|=\sqrt{1296}=36.$

5. Final answer

$\boxed{36}$

6. Comparison with stored correct answer

Stored correct answer = $36$ .

Our derived answer also gives $36$ , so they agree.

Permutations and Combinations question

Permutations and Combinations question

1. Basic allowed ordered pairs

2. Counting $n(Y)$

3. Counting $n(Z)$

4. Compute $n(Y)+n(Z)$

5. Final answer

6. Comparison with stored correct answer

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2. Counting $n(Y)$

3. Counting $n(Z)$

4. Compute $n(Y)+n(Z)$

5. Final answer

6. Comparison with stored correct answer

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