The total number of injective mappings from a set with m elements to a set with n elements, m ≤ n, is

(a) mn

(b) nm

(c) n!/(n-m)!

(d) n!


Let A= {a1, a2 …am}

B = {b1, b2 …bn} m ≤ n

Given f: A → B is injective mapping. 

For a1, there are n possible choices in B.

For a2, there are n-1 possible choices.

Similarly for am, there are n-m-1 choices. 

So the total number of injective mapping = nCm(m!)

= [n!/(n-m)!m!] m!

= n!/(n-m)!

Hence option c is the answer.

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