$$\because S={1,2,3,4,5,6}$$ and $$P(S) = \{ \phi ,\{ 1\} ,\{ 2\} ,....,\{ 1,2,3,4,5,6\} \} $$

$$f(n)$$ corresponding a set having m elements which belongs to P(S), should be a subset of $$f(n+1)$$, so $$f(n+1)$$ should be a subset of P(S) having at least $$m+1$$ elements.

Now, if f(1) has one element then f(2) has 3, f(3) has 3 and so on and f(6) has 6 elements. Total number of possible functions = 6! = 720 .... (1)

If f(1) has no elements (i.e. null set $$\phi$$) then

Each index number represents the number of elements in respective rows

Taking every series of arrow and counting number of such possible functions (sets)

$$ = {}^6{C_2} \times {}^4{C_1} \times {}^3{C_1} \times {}^2{C_1} + {}^6{C_1} \times {}^5{C_2} \times {}^3{C_1} \times {}^2{C_1} + {}^6{C_1} \times {}^5{C_1} \times {}^4{C_2} \times {}^2{C_1} + {}^6{C_1} \times {}^5{C_1} \times {}^4{C_1} \times {}^3{C_2} + {}^6{C_1} \times {}^5{C_1} \times {}^4{C_1} \times {}^3{C_1} \times {}^2{C_2} + {}^6{C_1} \times {}^5{C_1} \times {}^4{C_1} \times {}^3{C_1} \times {}^2{C_1}$$

$$ = 2520$$ ..........(2)

From (1) and (2) : Total number of functions

= 2520 + 720 = 3240