Basic Point :-
let no.of elements in A = p and no.of elements in B = q, then total functions = qp
let reduce your question into m=4 and n=5
How many functions can possible ?
4C0 * 45. ===> why multiplied by 4C0 is , we left 0 elements in A without mapping.
Partial Functions is atleast one element in A is left as without matching.
4C1 means 1 element in A left as without matching. ==> remaining (4-1) elements in A and 5 elements in B ===> 5(4-1) functions
4C2 means 2 element in A left as without matching. ==> remaining (4-2) elements in A and 5 elements in B ===> 5(4-2) functions
4C3 means 3 element in A left as without matching. ==> remaining (4-3) elements in A and 5 elements in B ===> 5(4-3) functions
4C4 means 4 element in A left as without matching. ==> remaining (4-4) elements in A and 5 elements in B ===> 5(4-4) functions
∴ 4C1 * 5(4-1) + 4C2 * 5(4-2) + 4C3 * 5(4-3) + 4C4 * 5(4-4) = 4C1 * 53 + 4C2 * 52 + 4C3 * 51 + 4C4 * 50 ===> are partial functions.
on generalization, it is like
No.of partial functions = mc1 . n(m-1) + mc2 . n(m-2) + mc3 . n(m-3) + .......+mci . n(m-i) + ......+mcm . n(m-m)
= mc1 . (1)1 . n(m-1) + mc2 . (1)2 . n(m-2) + mc3 . (1)3 . n(m-3) + .......+mci . (1)i . n(m-i) + ......+mcm . (1)m . n(m-m)
= mc0 . (1)0 . n(m) + mc1 . (1)1 . n(m-1) + mc2 . (1)2 . n(m-2) + mc3 . (1)3 . n(m-3) +....+mcm . (1)m . n(m-m) - mc0 . (1)0 . n(m)
= (1+n)m - mc0 . (1)0 . n(m)
= (1+n)m - n(m)