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(D) Let $S = \{1, 2, 3, 4, 5, 6\}$. We are looking for the number of one-one functions $f: S \rightarrow P(S)$ such that $f(1) \subset f(2) \subset f(

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$3240$

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(D) Let $S = \{1, 2, 3, 4, 5, 6\}$. We are looking for the number of one-one functions $f: S ightarrow P(S)$ such that $f(1) \subset f(2) \subset f(3) \subset f(4) \subset f(5) \subset f(6)$.This is equivalent to choosing a chain of $6$ distinct subsets of $S$ such that $A_1 \subset A_2 \subset A_3 \subset A_4 \subset A_5 \subset A_6$,where $A_i = f(i)$.Since $S$ has $6$ elements,the only way to have a chain of $6$ distinct subsets is to have the sizes of the subsets be $0, 1, 2, 3, 4, 5, 6$ in some order,but since they must be strictly increasing,the sizes must be $0, 1, 2, 3, 4, 5, 6$ in that specific order.However,we only have $6$ functions $f(1), \dots, f(6)$. Thus,we must choose $6$ subsets out of the $7$ possible sizes ($0$ to $6$).Let the sizes of the subsets be $s_1 $1$. $(0, 1, 2, 3, 4, 5)$: There are $\binom{6}{0} 	imes \binom{6-0}{1} 	imes \binom{5-0}{1} 	imes \binom{4-0}{1} 	imes \binom{3-0}{1} 	imes \binom{2-0}{1} = 1 	imes 6 	imes 5 	imes 4 	imes 3 	imes 2 = 720$ ways.$2$. $(0, 1, 2, 3, 4, 6)$: There are $\binom{6}{0} 	imes \binom{6}{1} 	imes \binom{5}{1} 	imes \binom{4}{1} 	imes \binom{3}{1} 	imes \binom{2}{2} = 1 	imes 6 	imes 5 	imes 4 	imes 3 	imes 1 = 360$ ways.$3$. $(0, 1, 2, 3, 5, 6)$: There are $\binom{6}{0} 	imes \binom{6}{1} 	imes \binom{5}{1} 	imes \binom{4}{1} 	imes \binom{3}{2} 	imes \binom{1}{1} = 1 	imes 6 	imes 5 	imes 4 	imes 3 	imes 1 = 360$ ways.$4$. $(0, 1, 2, 4, 5, 6)$: There are $\binom{6}{0} 	imes \binom{6}{1} 	imes \binom{5}{1} 	imes \binom{4}{2} 	imes \binom{2}{1} 	imes \binom{1}{1} = 1 	imes 6 	imes 5 	imes 6 	imes 2 	imes 1 = 360$ ways.$5$. $(0, 1, 3, 4, 5, 6)$: There are $\binom{6}{0} 	imes \binom{6}{1} 	imes \binom{5}{2} 	imes \binom{3}{1} 	imes \binom{2}{1} 	imes \binom{1}{1} = 1 	imes 6 	imes 10 	imes 3 	imes 2 	imes 1 = 360$ ways.$6$. $(0, 2, 3, 4, 5, 6)$: There are $\binom{6}{0} 	imes \binom{6}{2} 	imes \binom{4}{1} 	imes \binom{3}{1} 	imes \binom{2}{1} 	imes \binom{1}{1} = 1 	imes 15 	imes 4 	imes 3 	imes 2 	imes 1 = 360$ ways.$7$. $(1, 2, 3, 4, 5, 6)$: There are $\binom{6}{1} 	imes \binom{5}{1} 	imes \binom{4}{1} 	imes \binom{3}{1} 	imes \binom{2}{1} 	imes \binom{1}{1} = 6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 1 = 720$ ways.Total $= 720 + 360 + 360 + 360 + 360 + 360 + 720 = 3240$.

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

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