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)=6P(S)={φ | {1} | ...{6} | {1,2} | ... | |
{5,6} | ... | {1,2,3,4,5,6} |
}−64 elements case
−1f(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
=1080Case - 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
=720Case
−3f(6)=Sf(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
=360Case
−4f(6)=Sf(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
=360Case
−5f(6)=Sf(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
=360Case - 6
f(6)=Sf(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