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How many surjective functions exist from the set $A= \{1,2,3,4,5,6\}$ (Domain) to the set $B=\{w,x,y,z\}$ (Image).

I have no idea on how you do this. Any hints would be helpful.

I know that I can begin with $4^6$ functions and exclude the non surjective ones, but I don't know how you do this.

N. F. Taussig
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  • These are called Stirling numbers of the second kind...https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind – lulu Dec 05 '15 at 22:48
  • I didn't learn stirling numbers... –  Dec 05 '15 at 22:50
  • Or of http://math.stackexchange.com/questions/492433/if-a-30-and-b-20-find-the-number-of-surjective-functions-fa-to-b – Henry Dec 05 '15 at 22:51

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If you want to do this by inclusion-exclusion rather than Stirling numbers of the second kind then you will get $${4 \choose 4}\times 4^6 - {4 \choose 3}\times 3^6 + {4 \choose 2}\times 2^6 -{4 \choose 1}\times 1^6 +{4 \choose 0}\times 0^6.$$

Henry
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  • What formula did you use ? –  Dec 05 '15 at 23:25
  • @Astroman: inclusion-exclusion. Clearly $4^6$ counts some functions which are not surjective since they send $A$ to subsets of $B$ with three or fewer elements. So subtract off $3^6$ four times as there are four subsets of $B$ with three elements. But you have taken off the functions which send $A$ to subsets of $B$ with two or fewer elements too often, so add back $2^6$ six times as there are six subsets of $B$ with two elements. Then deal subsets of $B$ with one or zero elements in the same way... – Henry Dec 06 '15 at 00:25
  • But did you begin your sum at k=0 to 3 and substracted this sum from $4^6$ –  Dec 06 '15 at 00:28
  • @Astroman: I began my calculation at $4^6$ – Henry Dec 06 '15 at 00:46