Let S _ 1 = \ (i, j, k): i, j, k ∈ \ 1, 2, , 10\ \ S _ 2 = \ (i,...

${S}_{i} = \left\{\left(i, j, k\right): i, j, k \in \left\{1, 2, ..., 10\right\}\right\}$

${n}_{1} = 10 \times 10 \times 10 = 1000$

i.e., number of elements in ${S}_{1} = 1000$

${S}_{2} = \left\{\left(i, j\right): 1 \le i < j + 2 \le 10, i, j \in \left\{1, 2, ..., 10\right\}\right\}$

${n}_{2} = 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9$

$= 44$

${S}_{3} = \left\{\left(i, j, k, l\right): 1 \le i < j < k < l, i, j, k, l \in \left\{1, 2, ....10\right\}\right\}$

${n}_{3}{=}^{10}{C}_{4}$

$= \dfrac{10 \times 3 \times 8 \times 7 \times 6 !}{4 \times 3 \times 2 \times 1 \times 6 !} = 210$

i.e. number of elements in ${S}_{3} = 210$

${S}_{4} = \left\{\left(i, j, k, l\right): i, j, k\text{and}l\text{are distinct elements in}\left\{1, 2, ......., 10\right\}\right\}$

${n}_{4} = 10 \times 9 \times 8 \times 7$ or ${}^{10}{C}_{4} \times 4 !$

$= 5040$

$\dfrac{{n}_{4}}{12} = \dfrac{5040}{12} = 420$