IIT JEE , AIEEE Forum - Mathematics , Algebra

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As the functions to be formed are onto, the two elements in the co-domain set must have at least one pre-image.
For the first of these two elements, the number of ways of selecting a pre-image is n & for the second element is n-1
No. of ways of selecting one distinct pre-image is n(n-1).
The rest n-2 elements in the domain set can link to either of the two elements in the co-domain set. Basically the problem reduces to dividing this set of n-2 elements into two sets, each containing atleast one element. The number of ways to do this is n-3.

Hence the total number of ways to do this is n(n-1)(n-3)

This formula is valid only for n>3.
For n=1, no onto functions are possible.
For n =2, only one onto function is possible.
For n= 3, the no. of ways of selecting one preimage for the two elements is 3x2 = 6. For the remaining one element in the domain there are only two choices. Hence the total number of functions is 6 x 2 = 12. But out of these only 6 are distinct.

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