The correct option is B n!(n−m)!
Let A={a1,a2,a3.....am}
and B={b1,b2,b3.....bn} where m≤n
Given f:A→B be an injective mapping.
So, for a1∈A, there are n possible choices for f(a1)∈B.
For a2∈A, there are (n−1) possible choices for f(a2)∈B.
Similarly for am∈A, there are (n−m−1) choices for f(am)∈B
So, there are n(n−1)(n−2).....(n−m−1)=n!(n−m)! injective mapping from A to B.