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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

Option: 1

43


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

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No. of element 1=\{3\}
No. of element 2=\{(3 \mathrm{~K}+1),(3 \mathrm{k}+2)\}
                             \text { (3) }(3)=9

No. of element 3=\{3 \mathrm{k}, 3 \mathrm{k}+1,3 \mathrm{~K}+2\} \quad=$ (1) (3) $(3)=9=\{(3 \mathrm{k}+1),(3 \mathrm{k}+1),(3 \mathrm{k}+1)\}=1
=\{(3 \mathrm{~K}+2),(3 \mathrm{k}+2),(3 \mathrm{k}+2)\}=\frac{1}{11}

No. of element 4=\{3 \mathrm{k}, 3 \mathrm{k}+1,3 \mathrm{k}+1,3 \mathrm{k}+1\} \rightarrow 1
=\{3 \mathrm{k}, 3 \mathrm{k}+2,3 \mathrm{k}+2,3 \mathrm{k}+2\} \rightarrow 1
=(3 \mathrm{k}+1,3 \mathrm{k}+2,3 \mathrm{k}+2,3 \mathrm{k}+1\} \rightarrow{ }^{3} \mathrm{C}_{2} \times{ }^{3} \mathrm{C}_{2}=9

No. of element 5=9, no. of element 6=1, no. of element 7=1

\text { Total }=43 .

Posted by

Kuldeep Maurya

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