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← JEE Advanced 2021 · Paper 2
MathematicsMultiple correct+4 / -2Question 39 of 57

JEE Advanced 2021 · Paper 2, Mathematics question 1

Let
S1={(i,j,k):i,j,k∈{1,2,…,10}},S2={(i,j):1≤i<j+2≤10, i,j∈{1,2,…,10}},S3={(i,j,k,l):1≤i<j<k<l, i,j,k,l∈{1,2,…,10}}\begin{aligned}S_1 &= \{(i, j, k) : i, j, k \in \{1, 2, \ldots, 10\}\},\\ S_2 &= \{(i, j) : 1 \le i < j + 2 \le 10,\ i, j \in \{1, 2, \ldots, 10\}\},\\ S_3 &= \{(i, j, k, l) : 1 \le i < j < k < l,\ i, j, k, l \in \{1, 2, \ldots, 10\}\}\end{aligned}
and
S4={(i,j,k,l):i,j,k and l are distinct elements in {1,2,…,10}}.S_4 = \{(i, j, k, l) : i, j, k \text{ and } l \text{ are distinct elements in } \{1, 2, \ldots, 10\}\}.
If the total number of elements in the set SrS_r is nrn_r, r=1,2,3,4r = 1, 2, 3, 4, then which of the following statements is (are) TRUE?

Choose one or more options
Answer and worked solution

Official answer: A, B, D

Count each set.

  • n1=103=1000n_1=10^3=1000.
  • S2S_2: the condition i<j+2≤10i<j+2\le10 means j≤8j\le8 and 1≤i≤j+11\le i\le j+1. So n2=∑j=18(j+1)=44n_2=\sum_{j=1}^8(j+1)=44.
  • n3=(104)=210n_3=\binom{10}4=210, not 220.
  • n4=10⋅9⋅8⋅7=5040n_4=10\cdot9\cdot8\cdot7=5040, so n412=420\frac{n_4}{12}=420.

(A), (B), (D)

Topics:CombinationsPermutationsFundamental principle of counting
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