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  • Let Then the number of one-one functions <img alt="f: S \rightarrow

Let S = {1,2,3,4,5,6}.Then the number of one-one functions f: S \rightarrow P(S) where P(S) denote the power set of S, such that \mathrm{f}(\mathrm{n}) \subset \mathrm{f}(\mathrm{m}) where n> m

Option: 1

3240


Option: 2

__


Option: 3

__


Option: 4

__


Answers (1)

best_answer

Case – I
f(6) = S i.e. 1 option
f(5) = any 5 element subset A of S i.e.{ }^6 \mathrm{C}_5=6 \text { options } \\

f(4) = any 4 element subset B of A i.e.{ }^5 \mathrm{C}_4=5 \text { options } \\
f(3) = any 3 element subset C of B i.e.{ }^4 \mathrm{C}_3=4 \text { options } \\

f(2) = any 2 element subset D of C i.e.{ }^3 \mathrm{C}_2=3 \text { options }

f(1) = any 1 element subset E of D or empty subset i.e. 3 options

Total function = 6\times 5\times 4\times 3\times 2\times 3=1080

Case – II
f(6) = S
f(5) = any 4 element subset A of S i.e.{ }^6 \mathrm{C}_4=15 \text { options } \\

f(4) = any 3 element subset B of A i.e.{ }^4 \mathrm{C}_3=4 \text { options } \\

f(3) = any 2 element subset C of B i.e.{ }^3 \mathrm{C}_2=2 \text { options } \\

f(2) = any 1 element subset D of C i.e.{ }^2 \mathrm{C}_1=2 \text { options }

f(1) = empty subset i.e. 1 options
Total function = 15\times 4\times 3\times 2\times 1=360

Case – III
f(6) = S
f(5) = any 5 element subset A of S i.e.{ }^6 \mathrm{C}_5=6 \text { options } \\

f(4) = any 3 element subset B of A i.e.{ }^5 \mathrm{C}_3=10 \text { options } \\

f(3) = any 2 element subset C of B i.e.{ }^3 \mathrm{C}_2=3 \text { options } \\

f(2) = any 1 element subset D of C i.e.{ }^2 \mathrm{C}_1=2 \text { options }

f(1) = empty subset i.e. 1 options
Total function = 6\times 10\times 3\times 2\times 1=360

Case – IV
f(6) = S
f(5) = any 5 element subset A of S i.e.{ }^6 \mathrm{C}_5=6 \text { options } \\

f(4) = any 4 element subset B of A i.e.{ }^5 \mathrm{C}_4=5 \text { options } \\

f(3) = any 2 element subset C of B i.e.{ }^4 \mathrm{C}_2=6 \text { options } \\

f(2) = any 1 element subset D of C i.e.{ }^2 \mathrm{C}_1=2 \text { options }

f(1) = empty subset i.e. 1 options
Total function = 6\times 5\times 6\times 2\times 1=360

Case – V
f(6) = S
f(5) = any 5 element subset A of S i.e.{ }^6 \mathrm{C}_5=6 \text { options } \\

f(4) = any 4 element subset B of A i.e.{ }^5 \mathrm{C}_4=5 \text { options } \\

f(3) = any 3 element subset C of B i.e.{ }^4 \mathrm{C}_3=4 \text { options } \\

f(2) = any 1 element subset D of C i.e.{ }^3 \mathrm{C}_1=3 \text { options }

f(1) = empty subset i.e. 1 options
Total function = 6\times 5\times 4\times 3\times 1=360

Case – VI
f(6) = any 5 element subset A of S i.e.{ }^6 \mathrm{C}_5=6 \text { options } \\

f(5) = any 4 element subset B of A i.e.{ }^5 \mathrm{C}_4=5 \text { options } \\

f(4) = any 3 element subset C of B i.e.{ }^4 \mathrm{C}_3=4 \text { options } \\

f(3) = any 2 element subset D of C i.e.{ }^3 \mathrm{C}_2=3 \text { options } \\

f(2) = any 1 element subset E of D i.e.{ }^2 \mathrm{C}_1=2 \text { options }

f(1) = empty subset i.e. 1 options
Total function =6\times 5\times 4\times 3 \times 2\times 1=720

Total number of such functions ={ }^2 \mathrm{C}_1=2 \text { options }1080+(4 \times 360)+720=\mathbf{3 2 4 0}

 

 

 

 

 

 

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Sayak

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