The number of surjections f: [8] -> [5]...
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The number of surjections f: [8] -> [5]...

[From: ] [author: ] [Date: 12-08-16] [Hit: ]
this problem is a bit subtle and easy to mess up. Posting a wrong answer on this could have been most embarrassing on my part!......
Is it something like (5 choose 3) * 5! * 3! ?

I'd like to confirm this for an exam tomorrow.

My train of thought was 'Choose 3 from 5, to be doubled up on, and then arrange the first 5 matches, and finally to arrange the doubles.'

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We count this by inclusion-exclusion.
#(total number of mappings) - #(missing at least one element of [5]) + #(missing at least 2 elements of [5]) - #(missing at least 3 elements of [5]) + #(missing at least 4 elements of [5]) - #(missing all 5 elements of [5])

This simplifies to
5^8 - C(5, 1) * (5 - 1)^8 + C(5, 2) * (5 - 2)^8 - C(5, 3) * (5 - 3)^8 + C(5, 4) * (5 - 4)^8 - 0
= 5^8 - C(5, 1) * 4^8 + C(5, 2) * 3^8 - C(5, 3) * 2^8 + C(5, 4) * 1^8.

I hope this helps!

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Pardon the delay; as you see, this problem is a bit subtle and easy to mess up. Posting a wrong answer on this could have been most embarrassing on my part!

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