JEE Mains 30-Jan-2023 Shift 1 Solved Paper

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
‌n(S)=6
‌P(S)={

φ	{1}	...{6}	{1,2}	...	‌
{5,6}	...	{1,2,3,4,5,6}

}
‌‌‌−64‌ elements ‌
case −1
f(6)=S i.e. 1 option,
f(5)= any 5 element subset A of S i.e. 6 options,
f(4)= any 4 element subset B of A i.e. 5 options,
f(3)= any 3 element subset C of B i.e. 4 options,
f(2)= any 2 element subset D of C i.e. 3 options,
f(1)= any 1 element subset E of D or empty subset i.e. 3
options,
Total functions =1080
Case - 2
f(6)= any 5 element subset A of S i.e. 6 options,
f(5)= any 4 element subset B of A i.e. 5 options,
f' (4) = any 3 element subset C of B i.e. 4 options,
f(3)= any 2 element subset D of C i.e. 3 options,
f' (2)= any 1 element subset E of D i.e. 2 options,
f(1)= empty subset i.e. 1 option
Total functions =720
Case −3
f(6)=S
f(5)= any 4 element subset A of' S i.e. 15 options,
f(4)= any 3 element subset B of A i.e. 4 options,
f(3)= any 2 element subset C of B i.e. 3 options,
f(2)= any 1 element subset D of C i.e. 2 options,
f(1)= empty subset i.e. 1 option

Total functions =360
Case −4
f(6)=S
f(5)= any 5 element subset A of S i.e. 6 options,
f(4)= any 3 element subset B of A i.e. 10 options,
f(3)= any 2 element subset C of B i.e. 3 options,
f(2)= any 1 element subset D of C i.e. 2 options,
f(1)= empty subset i.e. 1 option
Total functions =360
Case −5
f(6)=S
f(5)= any 5 element subset A of S i.e. 6 options,
f(4)= any 4 element subset B of A i.e. 5 options,
f(3)= any 2 element subset C of B i.e. 6 options,
f(2)= any 1 element subset D of C i.e. 2 options,
f(1)= empty subset i.e. 1 option
Total functions =360
Case - 6
f(6)=S
f(5)= any 5 element subset A of S i.e. 6 options,
f(4)= any 4 element subset B of A i.e. 5 options,
f(3)= any 3 element subset C of B i.e. 4 options,
f(2)= any 1 element subset D of C i.e. 3 options,
f(1)= empty subset i.e. 1 option
Total functions =360
∴ Number of such functions =3240