Let S = {1, 2, 3, 4, 5, 6}. Then the number of one-one functions \(f : S → P(S)\), where \(P(S)\) denotes the power set of S, such that \(f(n) ⊂ f(m)\) where \(n < m,\) is:
Correct Answer: 3240
Solution and Explanation
We are given the set \(S = \{1, 2, 3, 4, 5, 6\}\). The number of elements in \(S\) is:
\[ n(S) = 6. \]
The power set \(P(S)\) contains all subsets of \(S\), including the empty set, and has:
\[ |P(S)| = 2^6 = 64 \text{ elements.} \]
Case Analysis for the One-One Functions
We need to count the one-one functions \(f : S \to P(S)\) such that \(f(n) \subset f(m)\) for \(n < m\).
Case 1
\(f(6) = S\) (1 option).
\(f(5) =\) any 5-element subset of \(S\) (6 options).
\(f(4) =\) any 4-element subset of \(f(5)\) (5 options).
\(f(3) =\) any 3-element subset of \(f(4)\) (4 options).
\(f(2) =\) any 2-element subset of \(f(3)\) (3 options).
\(f(1) =\) any 1-element subset of \(f(2)\) or the empty subset (3 options).
Total functions:
\[ 1 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 3 = 1080. \]
Case 2
\(f(6) =\) any 5-element subset of \(S\) (6 options).
\(f(5) =\) any 4-element subset of \(f(6)\) (5 options).
\(f(4) =\) any 3-element subset of \(f(5)\) (4 options).
\(f(3) =\) any 2-element subset of \(f(4)\) (3 options).
\(f(2) =\) any 1-element subset of \(f(3)\) (2 options).
\(f(1) =\) the empty subset (1 option).
Total functions:
\[ 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 720. \]
Case 3
\(f(6) = S\) (1 option).
\(f(5) =\) any 4-element subset of \(S\) (15 options).
\(f(4) =\) any 3-element subset of \(f(5)\) (4 options).
\(f(3) =\) any 2-element subset of \(f(4)\) (3 options).
\(f(2) =\) any 1-element subset of \(f(3)\) (2 options).
\(f(1) =\) the empty subset (1 option).
Total functions:
\[ 1 \cdot 15 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 360. \]
Cases 4, 5, and 6
Similarly, other configurations of the subsets give 360 functions each.
Total Number of Functions
Add the functions from all cases:
\[ 1080 + 720 + 360 + 360 + 360 + 360 = 3240. \]
Conclusion
The total number of such functions is:
\[ \boxed{3240}. \]
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