JEE MAIN - Mathematics (2023 - 25th January Morning Shift - No. 17)
Let S = {1, 2, 3, 5, 7, 10, 11}. The number of non-empty subsets of S that have the sum of all elements a multiple of 3, is _____________.
Answer
43
Explanation
Elements of the type $3 \mathrm{k}=3$
Elements of the type $3 \mathrm{k}+1=1,7,9$
Elements of the type $3 \mathrm{k}+2=2,5,11$
Subsets containing one element $S_1=1$
Subsets containing two elements
$$ S_2={ }^3 C_1 \times{ }^3 C_1=9 $$
Subsets containing three elements
$$ \mathrm{S}_3={ }^3 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1+1+1=11 $$
Subsets containing four elements
$$
\mathrm{S}_4={ }^3 \mathrm{C}_3+{ }^3 \mathrm{C}_3+{ }^3 \mathrm{C}_2 \times{ }^3 \mathrm{C}_2=11 $$
Subsets containing five elements
$$ \mathrm{S}_5={ }^3 \mathrm{C}_2 \times{ }^3 \mathrm{C}_2 \times 1=9 $$
Subsets containing six elements $\mathrm{S}_6=1$
Subsets containing seven elements $\mathrm{S}_7=1$
$$ \Rightarrow \text { sum }=43 $$
Elements of the type $3 \mathrm{k}+1=1,7,9$
Elements of the type $3 \mathrm{k}+2=2,5,11$
Subsets containing one element $S_1=1$
Subsets containing two elements
$$ S_2={ }^3 C_1 \times{ }^3 C_1=9 $$
Subsets containing three elements
$$ \mathrm{S}_3={ }^3 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1+1+1=11 $$
Subsets containing four elements
$$
\mathrm{S}_4={ }^3 \mathrm{C}_3+{ }^3 \mathrm{C}_3+{ }^3 \mathrm{C}_2 \times{ }^3 \mathrm{C}_2=11 $$
Subsets containing five elements
$$ \mathrm{S}_5={ }^3 \mathrm{C}_2 \times{ }^3 \mathrm{C}_2 \times 1=9 $$
Subsets containing six elements $\mathrm{S}_6=1$
Subsets containing seven elements $\mathrm{S}_7=1$
$$ \Rightarrow \text { sum }=43 $$
Comments (0)
