Question

If n(X) = ^mC₆, then the value of m is _______.

Answer 1

Correct Answer: 20

Condition Satisfaction 

To satisfy the condition \( |a - b| \geq 2 \), \( b \) must be at least 2 units away from \( a \). The possible pairs are:

• a = 1, b = 3, 4, 5, 6
• a = 2, b = 4, 5, 6
• a = 3, b = 1, 5, 6
• a = 4, b = 1, 2, 6
• a = 5, b = 1, 2, 3
• a = 6, b = 1, 2, 3, 4

Counting all valid pairs:

\[ \text{Total pairs} = 20 \]

• \( n(X) = 20 \)
• \( m = 20 \)

Answer 2

We need to find \(m\) such that \(n(X) = \binom{m}{6}\), where \(X\) is the set of all relations \(R\) from \(S\) to \(S\) with the following properties:

• \(|R| = 6\) (exactly 6 ordered pairs in the relation)
• For each \((a, b) \in R\), \(|a - b| \geq 2\)

Given:

• Set \(S = \{1, 2, 3, 4, 5, 6\}\)
• Relations \(R \subseteq S \times S\)
• Relations \(R\) have size 6
• Condition on pairs: \(|a - b| \geq 2\)

Step 1: Count the total number of ordered pairs \((a, b)\) in \(S \times S\) satisfying \(|a - b| \geq 2\).

For each \(a \in S\), find all \(b \in S\) with \(|a - b| \geq 2\).

\(a\)	Values of \(b\) with \(|a - b| \geq 2\)	Count
1	3, 4, 5, 6	4
2	4, 5, 6	3
3	1, 5, 6	3
4	1, 2, 6	3
5	1, 2, 3	3
6	1, 2, 3, 4	4

Total number of such pairs:

\[ 4 + 3 + 3 + 3 + 3 + 4 = 20 \]

Step 2: The set \(X\) consists of all subsets of these 20 pairs with exactly 6 elements.
Thus, \[ n(X) = \binom{20}{6} \] Given, \[ n(X) = \binom{m}{6} \] Comparing, \[ m = 20 \]

Final Answer:
\[ \boxed{20} \]

Answer 3

Let $S = \{1, 2, 3, 4, 5, 6\}$. Let $X$ be the set of all relations $R$ from $S$ to $S$ such that $|R|=6$ and for each $(a, b) \in R$, $|a - b| \ge 2$.

$Y = \{R \in X : The \ range \ of \ R \ has \ exactly \ one \ element\}$.
$Z = \{R \in X : R \ is \ a \ function \ from \ S \ to \ S\}$.

For $Y$:
The range of $R$ has exactly one element, say $b \in S$. So $R = \{(a_1, b), (a_2, b), (a_3, b), (a_4, b), (a_5, b), (a_6, b)\}$. We need $|a_i - b| \ge 2$ for all $i = 1, 2, ..., 6$.
So $a_i \in S$ and $|a_i - b| \ge 2$. Also, $a_i$ must be distinct.

• If $b = 1$, then $a_i \in \{3, 4, 5, 6\}$. We need to choose 6 distinct elements from this set, which is impossible.
• If $b = 2$, then $a_i \in \{4, 5, 6\}$. We need to choose 6 distinct elements from this set, which is impossible.
• If $b = 3$, then $a_i \in \{1, 5, 6\}$. We need to choose 6 distinct elements from this set, which is impossible.
• If $b = 4$, then $a_i \in \{1, 2, 6\}$. We need to choose 6 distinct elements from this set, which is impossible.
• If $b = 5$, then $a_i \in \{1, 2, 3\}$. We need to choose 6 distinct elements from this set, which is impossible.
• If $b = 6$, then $a_i \in \{1, 2, 3, 4\}$. We need to choose 6 distinct elements from this set, which is impossible.

$n(Y) = 0$.

For $Z$:
$R$ is a function from $S$ to $S$. So $R = \{(1, b_1), (2, b_2), (3, b_3), (4, b_4), (5, b_5), (6, b_6)\}$, where $b_i \in S$.
We need $|i - b_i| \ge 2$ for all $i = 1, 2, ..., 6$.

• $i = 1$: $b_1 \in \{3, 4, 5, 6\}$. 4 choices.
• $i = 2$: $b_2 \in \{4, 5, 6\}$. 3 choices.
• $i = 3$: $b_3 \in \{1, 5, 6\}$. 3 choices.
• $i = 4$: $b_4 \in \{1, 2, 6\}$. 3 choices.
• $i = 5$: $b_5 \in \{1, 2, 3\}$. 3 choices.
• $i = 6$: $b_6 \in \{1, 2, 3, 4\}$. 4 choices.

So $n(Z) = 4 \times 3 \times 3 \times 3 \times 3 \times 4 = 36 \times 9 \times 4 = 36 \times 36 = 1296$.
$n(Y) + n(Z) = 0 + 1296 = 1296 = k^2$.
$|k| = \sqrt{1296} = \sqrt{36^2} = 36$.

Final Answer:
The final answer is $\boxed{36}$.

Answer 4

Given:

• \(Y = \{ R \in X : \text{range of } R \text{ has exactly one element} \}\)
• \(Z = \{ R \in X : R \text{ is a function from } S \to S \}\)
• Need to find \(|k|\) where \(k^2 = n(Y) + n(Z)\).

Step 1: Find \(n(Y)\)
- Each \(R \in Y\) has 6 ordered pairs, all with the same second component (range has exactly one element).
- For fixed \(b \in S\), choose pairs \((a, b)\) where \(a \in S\) and \(|a - b| \geq 2\).
- Since \(|a-b| \geq 2\), we must check for each \(b\) how many \(a\)'s satisfy it, but in \(Y\), the pairs are \((a,b)\) with the same \(b\), so we fix \(b\) and choose the 6 pairs from allowed \(a\)'s.
- However, the relation has exactly 6 elements. Since domain \(S\) has size 6, to have exactly 6 elements all with the same \(b\), we must select all \(a \in S\) such that \(|a-b| \geq 2\) and the total number of such \(a\) equals 6.
- Check if for some \(b\), there are exactly 6 values of \(a\) with \(|a-b| \geq 2\).

\(b\)	Allowed \(a\) with \(|a - b| \geq 2\)	Count
1	3,4,5,6	4
2	4,5,6	3
3	1,5,6	3
4	1,2,6	3
5	1,2,3	3
6	1,2,3,4	4

No \(b\) has 6 \(a\)'s satisfying \(|a-b| \geq 2\). Therefore, \(n(Y) = 0\).

Step 2: Find \(n(Z)\)
- \(Z\) consists of relations \(R\) that are functions from \(S\) to \(S\), with the restriction \(|a - b| \geq 2\) for \((a,b) \in R\), and \(|R|=6\) (one output for each \(a\)).
- For each \(a\), we choose exactly one \(b\) with \(|a-b| \geq 2\).
- The number of such functions is the product over \(a\) of the number of allowed \(b\)'s for that \(a\).

From earlier, number of \(b\)'s for each \(a\) with \(|a-b| \geq 2\):

\(a\)	Allowed \(b\) with \(|a-b| \geq 2\)	Count
1	3,4,5,6	4
2	4,5,6	3
3	1,5,6	3
4	1,2,6	3
5	1,2,3	3
6	1,2,3,4	4

Therefore, \[ n(Z) = 4 \times 3 \times 3 \times 3 \times 3 \times 4 = 4^2 \times 3^4 = 16 \times 81 = 1296 \]

Step 3: Calculate \(|k|\)
Since \(n(Y) = 0\), \[ n(Y) + n(Z) = 0 + 1296 = 1296 \] Note that, \[ 1296 = 36^2 \] So, \[ |k| = 36 \]

Final Answer:
\[ \boxed{36} \]