LetS1={(i,j,k):i,j,k∈{1,2,….,10}}S2={(i,j):1≤i

(A) n 1 = 10 × 10 × 10 = 1000 (B) As per given condition 1 ≤ i < j + 2 ≤ 10 ⇒ j ≤ 8 and i ≥ 1 For i = 1 , 2 , j = 1 , 2 , 3 , … … , 8 → ( 8 + 8 ) possibilities For i = 3 , j = 2 , 3 , … … , 8 → 7 possibilities i = 9 , j = 1 → 1 possibility So n 2 = ( 1 + 2 + 3 + … … + 8 ) + 8 = 44 (C) n 3 = 10 C 4 (Choose any four) = 210 (D) n 4 = 10 C 4 ⋅ 4 ! = ( 210 ) ( 24 ) ⇒ n 4 12 = 420

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Let
$\begin{array}{l} S_{1} = {(i, j, k) : i, j, k \in {1, 2, \dots ., 10}} \\ S_{2} = {(i, j) : 1 \leq i < j + 2 \leq 10, i, j \in {1, 2, \dots \dots, 10}} \\ S_{3} = {(i, j, k, ℓ) : 1 \leq i < j < k < ℓ, i, j, k, ℓ \in {1, 2, \dots ., 10}} \end{array}$
and
$S_{4} = {(i, j, k, ℓ) : i, j, k and ℓ$ are distinct elements in {1, 2, …..,10}}.
If the total number of elements in the set $S_{r} is n_{r}$ , r = 1, 2, 3, 4, then which of the
following statements is (are) TRUE ? ([.] denotes GIF)

see full answer

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$[\frac{n_{2}}{11}] = 4$

$[\frac{n_{1}}{50}] = 20$

$[\frac{n_{4}}{21}] = 420$

$[\frac{n_{3}}{21}] = 10$

answer is A, B, C.

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Detailed Solution

$(A) n_{1} = 10 \times 10 \times 10 = 1000$
(B) As per given condition $1 \leq i < j + 2 \leq 10 \Rightarrow j \leq 8 and i \geq 1$
For $i = 1, 2, j = 1, 2, 3, \dots \dots, 8 \to (8 + 8)$ possibilities
For $i = 3, j = 2, 3, \dots \dots, 8 \to 7$ possibilities
$i = 9, j = 1 \to 1$ possibility
$So n_{2} = (1 + 2 + 3 + \dots \dots + 8) + 8 = 44$
$(C) n_{3} =^{10} C_{4}$ (Choose any four)
$= 210 \begin{array}{l} (D) n_{4} =^{10} C_{4} \cdot 4! = (210) (24) \\ \Rightarrow \frac{n_{4}}{12} = 420 \end{array}$