Number of Elements in Defined Sets of Tuples and Combinations

To find the correct statements, we will evaluate the total number of elements $n_r$ for each set $S_r$ ($r = 1, 2, 3, 4$).

1. Evaluation of $n_1$ (Set $S_1$):

The set $S_1$ is defined as: $S_1 = \{(i,j,k) : i, j, k \in \{1, 2, \dots, 10\}\}$

Since each of $i$, $j$, and $k$ can independently take any of the $10$ integer values from $\{1, 2, \dots, 10\}$, the total number of elements $n_1$ is given by: $n_1 = 10 \times 10 \times 10 = 1000$

Thus, Option (A) is TRUE.

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2. Evaluation of $n_2$ (Set $S_2$):

The set $S_2$ is defined as: $S_2 = \{(i,j) : 1 \le i < j + 2 \le 10, \text{ with } i, j \in \{1, 2, \dots, 10\}\}$

From the inequality $j + 2 \le 10$, we get: $j \le 8$ Since $j \in \{1, 2, \dots, 10\}$, the possible values for $j$ are $j \in \{1, 2, \dots, 8\}$.

For a fixed value of $j$, the inequality $1 \le i < j + 2$ implies: $1 \le i \le j + 1$ Thus, for a given $j$, $i$ can take $(j + 1)$ distinct values.

Summing over all valid values of $j \in \{1, 2, \dots, 8\}$, we get: $n_2 = \sum_{j=1}^{8} (j + 1) = 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9$ $n_2 = \frac{8 \times (2 + 9)}{2} = 44$

Thus, Option (B) is TRUE.

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3. Evaluation of $n_3$ (Set $S_3$):

The set $S_3$ is defined as: $S_3 = \{(i,j,k,l) : 1 \le i < j < k < l, \text{ with } i, j, k, l \in \{1, 2, \dots, 10\}\}$

This represents choosing $4$ distinct elements from the set $\{1, 2, \dots, 10\}$ and arranging them in strict ascending order, which can be done in only $1$ way for any chosen combination.

Thus, the total number of elements $n_3$ is: $n_3 = \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$

Since $n_3 = 210 \ne 220$, Option (C) is FALSE.

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4. Evaluation of $n_4$ (Set $S_4$):

The set $S_4$ is defined as: $S_4 = \{(i,j,k,l) : i, j, k \text{ and } l \text{ are distinct elements from } \{1, 2, \dots, 10\}\}$

The total number of ordered $4$-tuples of distinct elements chosen from $10$ available numbers is: $n_4 = {^{10}P_4} = 10 \times 9 \times 8 \times 7 = 5040$

Now, calculating $\frac{n_4}{12}$: $\frac{n_4}{12} = \frac{5040}{12} = 420$

Thus, Option (D) is TRUE.

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Conclusion:

The correct statements are (A), (B), and (D).