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Can someone show the derivation of number of one-one functions(f: A->B) is nPm where |A|=m and |B|=n?

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Your function is From A to B and  |A|=m and |B|=n

A to B is a Function means, Element from A should be mapped to only one element of B, it may be any element in B

A to B is a One-to-One Function means, Element from A should be mapped to only one element of B, it may be any element in B and  Element from B should be mapped to only one element of A, it may be any element in A

there are m elements in A, let name them as a1,a2,a3,a4,a5,a6....am and  there are n elements in B, let name them as b1,b2,b3,b4,b5,b6....bn

a1 can mapped to any element of B ====> n choices for mapping a1 ( Why n ? due to there are n elements in B)

a2 can mapped to any element of B except one element which is already mapped to a1 ===> n-1  choices for mapping a2

a3 can mapped to any element of B except two elements which is already mapped to a1 and a2 ===> n-2  choices for mapping a3

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am can mapped to any element of B except m-1 elements of B which is already mapped to a1 and a2,...am-1 ===> n-(m-1)  choices for mapping am

 

We have Multiply them = (n) . (n-1) . (n-2) ..... ( n- (m-1) ) =  $ \frac{ (n) . (n-1) . (n-2) ..... ( n- (m-1) ) {\color{Blue} (n-m)! }}{\color{Blue} (n-m)! } $ = $ \frac{n!}{(n-m)!}$ = nPm

 

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