QJEEM23Ma6a2b: JEE Main 2023 Mathematics - Class 11 JEEM23Ma6a2b Solution: Let . Then the number of one-one functions , where denote the…

$\because S={1,2,3,4,5,6}$ 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